Kuro and Walking Route CodeForces - 979C （树上DFS）

Kuro is living in a country called Uberland, consisting of nn towns, numbered from 11to nn, and n−1n−1 bidirectional roads connecting these towns. It is possible to reach each town from any other. Each road connects two towns aa and bb. Kuro loves walking and he is planning to take a walking marathon, in which he will choose a pair of towns (u,v)(u,v) (u≠vu≠v) and walk from uu using the shortest path to vv (note that (u,v)(u,v) is considered to be different from (v,u)(v,u)).

Oddly, there are 2 special towns in Uberland named Flowrisa (denoted with the index xx) and Beetopia (denoted with the index yy). Flowrisa is a town where there are many strong-scent flowers, and Beetopia is another town where many bees live. In particular, Kuro will avoid any pair of towns (u,v)(u,v) if on the path from uu to vv, he reaches Beetopia after he reached Flowrisa, since the bees will be attracted with the flower smell on Kuro’s body and sting him.

Kuro wants to know how many pair of city (u,v)(u,v) he can take as his route. Since he’s not really bright, he asked you to help him with this problem.

Input

The first line contains three integers nn, xx and yy (1≤n≤3⋅1051≤n≤3⋅105, 1≤x,y≤n1≤x,y≤n, x≠yx≠y) - the number of towns, index of the town Flowrisa and index of the town Beetopia, respectively.

n−1n−1 lines follow, each line contains two integers aa and bb (1≤a,b≤n1≤a,b≤n, a≠ba≠b), describes a road connecting two towns aa and bb.

It is guaranteed that from each town, we can reach every other town in the city using the given roads. That is, the given map of towns and roads is a tree.

Output

A single integer resembles the number of pair of towns (u,v)(u,v) that Kuro can use as his walking route.

Examples

Input

3 1 3
1 2
2 3

Output

5

Input

3 1 3
1 2
1 3

Output

4

Note

On the first example, Kuro can choose these pairs:

• (1,2)(1,2): his route would be 1→21→2,
• (2,3)(2,3): his route would be 2→32→3,
• (3,2)(3,2): his route would be 3→23→2,
• (2,1)(2,1): his route would be 2→12→1,
• (3,1)(3,1): his route would be 3→2→13→2→1.

Kuro can't choose pair (1,3)(1,3) since his walking route would be 1→2→31→2→3, in which Kuro visits town 11 (Flowrisa) and then visits town 33 (Beetopia), which is not allowed (note that pair (3,1)(3,1) is still allowed because although Kuro visited Flowrisa and Beetopia, he did not visit them in that order).

On the second example, Kuro can choose the following pairs:

• (1,2)(1,2): his route would be 1→21→2,
• (2,1)(2,1): his route would be 2→12→1,
• (3,2)(3,2): his route would be 3→1→23→1→2,
• (3,1)(3,1): his route would be 3→13→1.

题意：

给你一个含有N 个节点的树，

并且有两个节点比较特殊分别是x和y，

特殊之处是不能有一个路径是先走过x后又走过y。因为那样咱们题目的主公人会被蜜蜂叮咬。

求还剩多少个简单路径可以走？

思路：

一个图上所以简单路径的个数是 n*(n-1) .

而不能走的路径是先走过x后又走过y。

我们想一下会有哪些简单路径是先经过x后经过y。

我们看图，我画的一个不好看的树，

如果1是x，3是y，那么1和它左边连着的几个节点不能去3以及3右边的几个节点。

那么我们只需要由x节点在树上dfs时遇到y节点不继续搜，那么没访问到的节点的个数就是x不能去访问的y和y的几个节点。

反过来用y节点去dfs，可以得到x和x的那几个节点计数为numx吧，

那么numx*numy就是我们不能走的简单路径，总路径数减去不能走的就是我们要的答案数。

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 rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#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 db(x) cout<<"== [ "<<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=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
std::vector<int> son[maxn];
int vis1[maxn];
int vis2[maxn];
int bid;
int fid;
void dfs1(int id,int pre)
{
    vis1[id]=;
    for(auto x:son[id])
    {
        if(x!=pre&&x!=fid)
        {
            dfs1(x,id);
        }
    }
}
void dfs2(int id,int pre)
{
    vis2[id]=;
    for(auto x:son[id])
    {
        if(x!=pre&&x!=bid)
        {
            dfs2(x,id);
        }
    }
}
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>n>>fid>>bid;
    int a,b;
    repd(i,,n)
    {
        cin>>a>>b;
        son[a].pb(b);
        son[b].pb(a);
    }
    dfs1(bid,-);
    dfs2(fid,-);
    ll num1=0ll;
    ll num2=0ll;
    repd(i,,n)
    {
        if(!vis1[i])
        {
            num1++;
        }
        if(!vis2[i])
        {
            num2++;
        }
    }
    ll ans=1ll*n*(n-1ll);
    ans-=num1*num2;
    cout<<ans<<endl;

    return ;
}

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