Kay and Snowflake CodeForces - 686D (树的重心性质)

After the piece of a devilish mirror hit the Kay's eye, he is no longer interested in the beauty of the roses. Now he likes to watch snowflakes.

Once upon a time, he found a huge snowflake that has a form of the tree (connected acyclic graph) consisting of n nodes. The root of tree has index 1. Kay is very interested in the structure of this tree.

After doing some research he formed q queries he is interested in. The i-th query asks to find a centroid of the subtree of the node vi. Your goal is to answer all queries.

Subtree of a node is a part of tree consisting of this node and all it's descendants (direct or not). In other words, subtree of node v is formed by nodes u, such that node v is present on the path from u to root.

Centroid of a tree (or a subtree) is a node, such that if we erase it from the tree, the maximum size of the connected component will be at least two times smaller than the size of the initial tree (or a subtree).

Input

The first line of the input contains two integers n and q (2 ≤ n ≤ 300 000, 1 ≤ q ≤ 300 000) — the size of the initial tree and the number of queries respectively.

The second line contains n - 1 integer p2, p3, ..., pn (1 ≤ pi ≤ n) — the indices of the parents of the nodes from 2 to n. Node 1 is a root of the tree. It's guaranteed that pi define a correct tree.

Each of the following q lines contain a single integer vi (1 ≤ vi ≤ n) — the index of the node, that define the subtree, for which we want to find a centroid.

Output

For each query print the index of a centroid of the corresponding subtree. If there are many suitable nodes, print any of them. It's guaranteed, that each subtree has at least one centroid.

Example

Input

7 4

1 1 3 3 5 3

1

2

3

5

Output

3

2

3

6

Note

The first query asks for a centroid of the whole tree — this is node 3. If we delete node 3 the tree will split in four components, two of size 1 and two of size 2.

The subtree of the second node consists of this node only, so the answer is 2.

Node 3 is centroid of its own subtree.

The centroids of the subtree of the node 5 are nodes 5 and 6 — both answers are considered correct.

题意：

给你一个含有n个节点的树，并且给你q个询问，每一个询问x，问以x为根的子树中重心是哪个节点？

思路：

利用重心的2个性质：

1、对于一棵树来说，删去该树的重心后，所有的子树的大小不会超过原树大小的二分之一

2、两棵树合并后其新的重心在原两个重心的路径中。

那么我们在dfs树的过程中维护cntson[i]代表以i节点为根的子树大小，

然后在利用cntson可以找到一个子树的重心，在一个节点与其儿子节点为根的子树合并时，在当前维护的重心和儿子节点为根的子树的重心的路径中同样通过性质1来更新出当前的重心即可。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 300010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
std::vector<int> son[maxn];
int cntson[maxn];
int q;
int ans[maxn];
int f[maxn];
void dfs(int x, int pre)
{
    cntson[x] = 1;
    ans[x] = x;
    for (auto y : son[x])
    {
        if (y != pre)
        {
            dfs(y, x);
            cntson[x] += cntson[y];
        }
    }
    for (auto y : son[x])
    {
        if (y != pre)
        {
            if (cntson[y] * 2 > cntson[x])
            {
                ans[x] = ans[y];
            }
        }
    }

    while ((cntson[x] - cntson[ans[x]]) * 2 > cntson[x])
    {
        ans[x] = f[ans[x]];
    }

}
// int num;

int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gg(n);
    gg(q);
    repd(i, 2, n)
    {
        int x;
        gg(x);
        son[x].push_back(i);
        f[i] = x;
    }
    dfs(1, 1);
    int x;
    repd(i, 1, q)
    {
        gg(x);
        printf("%d\n", ans[x] );
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}