CodeForces 1062E Company

Description

The company $X$ has $n$ employees numbered from $1$ through $n$. Each employee $u$ has a direct boss $p_u$ $(1 \le p_u \le n)$, except for the employee $1$ who has no boss. It is guaranteed, that values $p_i$ form a tree. Employee $u$ is said to be in charge of employee $v$ if $u$ is the direct boss of $v$ or there is an employee $w$ such that $w$ is in charge of $v$ and $u$ is the direct boss of $w$. Also, any employee is considered to be in charge of himself.

In addition, for each employee $u$ we define it's level $lv(u)$ as follow:

$lv(1)=0$
$lv(u)=lv(p_u)+1$ for $u \neq 1$

In the near future, there are $q$ possible plans for the company to operate. The $i$-th plan consists of two integers $l_i$ and $r_i$, meaning that all the employees in the range $[l_i,r_i]$, and only they, are involved in this plan. To operate the plan smoothly, there must be a project manager who is an employee in charge of all the involved employees. To be precise, if an employee $u$ is chosen as the project manager for the $i$-th plan then for every employee $v \in [l_i,r_i]$, $u$ must be in charge of $v$. Note, that $u$ is not necessary in the range $[l_i,r_i]$. Also, $u$ is always chosen in such a way that $lv(u)$ is as large as possible (the higher the level is, the lower the salary that the company has to pay the employee).

Before any plan is operated, the company has JATC take a look at their plans. After a glance, he tells the company that for every plan, it's possible to reduce the number of the involved employees exactly by one without affecting the plan. Being greedy, the company asks JATC which employee they should kick out of the plan so that the level of the project manager required is as large as possible. JATC has already figured out the answer and challenges you to do the same.

Input

The first line contains two integers $n$ and $q$ $(2 \le n \le 100000, 1 \le q \le 100000)$ — the number of employees and the number of plans, respectively.

The second line contains $n−1$ integers $p_2,p_3,…,p_n(1≤p_i≤n)$ meaning $p_{i}$ is the direct boss of employee $i$.

It is guaranteed, that values $p_{i}$ form a directed tree with the root of $1$.

Each of the following $q$ lines contains two integers $l_i$ and $r_i$ $(1 \le l_i < r_i \le n)$ — the range of the employees, involved in the corresponding plan.

Output

Print $q$ lines, each containing two integers — the number of the employee which should be kicked from the corresponding plan and the maximum possible level of the project manager in that case.

If there are more than one way to choose that employee, print any of them.

Example

Input

11 5

1 1 3 3 3 4 2 7 7 6

4 6

4 8

1 11

9 11

8 11

Output

Note

In the example:

In the first query, we can choose whether $4$ or $5$ or $6$ and the project manager will be $3$.

In the second query, if we choose any employee other than the employee $8$, the project manager will be $1$. If we choose $8$, the project manager will be $3$. Since $lv(3)=1 \gt lv(1)=0$, choosing $8$ is the best strategy.

In the third query, no matter how we choose the employee, the project manager will always be $1$.

In the fourth query, if we choose $9$ or $10$ then the project manager will be $3$. If we choose $11$ then the project manager will be $7$. Since $lv(7)=3 \gt lv(3)=1$, we choose $11$ as the answer.

Solution

题意：给一棵树，$n$个点，$q$次询问，每次询问给定一个区间$[l, r]$，要求忽略掉$[l, r]$中的一个点，使得剩下的$r - l $个点的LCA的深度最大，问应该忽略哪个点，忽略后的最大深度是多少。

首先求一次DFS序，对于任意点$u$，其DFS序记为$order[u]$。给定区间$[l, r]$，设其中DFS序最大和最小的点分别为$u$和$v$，则$LCA[l, r]$就是$LCA(u, v)$。我们可以简单证明一下，不妨设$r = LCA(u, v)$，点$x$不属于以$r$为根的子树（记作$SubTree(r)$）当且仅当$order[x]$满足以下两种情况中的一种：

$order[x] \lt order[r]$，即$x$在$r$之前被访问
$order[x] > order[i], \forall i \in SubTree(r)$，即$x$在$SubTree(r)$之后才被访问

显然，$[l, r]$中的任何一个点都不满足上述两个条件，所以$[l, r]$中的每个点都属于以$r$为根的子树，所以它们的LCA就是$r$。

回到我们的问题，对于每次询问，给定$[l, r]$，我们先求出其中DFS序最大、最小的点$u, v$以及它们的LCA $r$。显然，忽略$u$和$v$之外的节点对并不会改变LCA；如果忽略$u$，那么新的LCA就是$LCA[l, u-1]$和$LCA[u + 1, r]$的LCA，我们称之为$r_1$；同理，忽略$v$也可以得到一个新的LCA，我们称之为$r_2$。选择$r, r_1, r_2$中深度最大的点，我们就得到了答案。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 100011;

const int maxp = 18;

vector<int> w[maxn];

int idx, dfsod[maxn], invdfsod[maxn];

int fa[maxn][maxp], dep[maxn];

void bfs(int root) {

  queue<int> que;

  dep[root] = 0;

  fa[root][0] = root;

  que.push(root);

  while (!que.empty()) {

    int u = que.front();

    que.pop();

    for (int i = 1; i < maxp; i++)

      fa[u][i] = fa[fa[u][i - 1]][i - 1];

    for (int v : w[u]) {

      if (v == fa[u][0])  continue;

      dep[v] = dep[u] + 1;

      fa[v][0] = u;

      que.push(v);

    }

  }

}

int lca(int u, int v) {

  if (dep[u] > dep[v])

    swap(u, v);

  for (int gap = dep[v] - dep[u], i = 0; gap; gap >>= 1, i++) {

    if (gap & 1)

      v = fa[v][i];

  }

  if (u == v) return u;

  for (int i = maxp - 1; i >= 0; i--) {

    if (fa[u][i] == fa[v][i])

      continue;

    u = fa[u][i], v = fa[v][i];

  }

  return fa[u][0];

}

void dfs(int u, int pre = -1) {

  dfsod[u] = ++idx;

  invdfsod[idx] = u;

  for (int v : w[u]) {

    if (v == pre) continue;

    dfs(v, u);

  }

}

struct node {

  int l, r, mx, mn;

} seg[maxn << 2];

void pushup(int x) {

  seg[x].mx = max(seg[x << 1].mx, seg[x << 1 | 1].mx);

  seg[x].mn = min(seg[x << 1].mn, seg[x << 1 | 1].mn);

}

void build(int x, int l, int r) {

  seg[x].l = l, seg[x].r = r;

  if (l == r) {

    seg[x].mx = seg[x].mn = dfsod[l];

    return;

  }

  int m = (l + r) >> 1;

  build(x << 1, l, m);

  build(x << 1 | 1, m + 1, r);

  pushup(x);

}

pair<int, int> query(int x, int l, int r) {

  int L = seg[x].l, R = seg[x].r;

  if (l <= L && r >= R)

    return make_pair(seg[x].mn, seg[x].mx);

  int m = (L + R) >> 1;

  int mx = 0, mn = 1 << 30;

  if (l <= m) {

    auto v = query(x << 1, l, r);

    mn = min(mn, v.first);

    mx = max(mx, v.second);

  }

  if (r > m) {

    auto v = query(x << 1 | 1, l, r);

    mn = min(mn, v.first);

    mx = max(mx, v.second);

  }

  return make_pair(mn, mx);

}

// 区间[l, r]的LCA

int getlca(int l, int r) {

  if (l > r)  return -1;

  auto x = query(1, l, r);

  int u = invdfsod[x.first], v = invdfsod[x.second];

  return lca(u, v);

}

// 忽略u后，区间[l, r]的LCA

int getlca(int l, int r, int u) {

  int a = getlca(l, u - 1), b = getlca(u + 1, r);

  if (a == -1) return b;

  if (b == -1) return a;

  return lca(a, b);

}

int main() {

  int n, q;

  scanf("%d%d", &n, &q);

  for (int i = 2; i <= n; ++i) {

    int x;  scanf("%d", &x);

    w[x].push_back(i);

    w[i].push_back(x);

  }

  bfs(1);

  dfs(1);

  build(1, 1, n);

  dep[0] = -1;

  while (q--) {

    int l, r;

    scanf("%d%d", &l, &r);

    auto x = query(1, l, r);

    int u = invdfsod[x.first], v = invdfsod[x.second];

    int c = lca(u, v), a = getlca(l, r, u), b = getlca(l, r, v);

    int mx = max(dep[c], max(dep[a], dep[b])), y;

    if (mx == dep[c]) y = l;

    else if (mx == dep[a])  y = u;

    else  y = v;

    printf("%d %d\n", y, mx);

  }

  return 0;

}