C. Timofey and a tree----思维题

C. Timofey and a tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Each New Year Timofey and his friends cut down a tree of n vertices and bring it home. After that they paint all the n its vertices, so that the i-th vertex gets color ci.

Now it's time for Timofey birthday, and his mother asked him to remove the tree. Timofey removes the tree in the following way: he takes some vertex in hands, while all the other vertices move down so that the tree becomes rooted at the chosen vertex. After that Timofey brings the tree to a trash can.

Timofey doesn't like it when many colors are mixing together. A subtree annoys him if there are vertices of different color in it. Timofey wants to find a vertex which he should take in hands so that there are no subtrees that annoy him. He doesn't consider the whole tree as a subtree since he can't see the color of the root vertex.

A subtree of some vertex is a subgraph containing that vertex and all its descendants.

Your task is to determine if there is a vertex, taking which in hands Timofey wouldn't be annoyed.

Input

The first line contains single integer n (2 ≤ n ≤ 105) — the number of vertices in the tree.

Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting there is an edge between vertices u and v. It is guaranteed that the given graph is a tree.

The next line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 105), denoting the colors of the vertices.

Output

Print "NO" in a single line, if Timofey can't take the tree in such a way that it doesn't annoy him.

Otherwise print "YES" in the first line. In the second line print the index of the vertex which Timofey should take in hands. If there are multiple answers, print any of them.

Examples

input

41 22 33 41 2 1 1

output

YES2

input

31 22 31 2 3

output

YES2

input

41 22 33 41 2 1 2

output

NO

题目链接：http://codeforces.com/contest/764/problem/C

一个非常好的思维题。

题意是说有没有一个点，去掉之后子图的颜色一样，如果有，就输出一个，如果有多个，就输出任意一个。

我们把这个题的边和点分开考虑，如果一个图中有不同的颜色，则必有一条边的两端不是同一种颜色，我们称这种边为特殊边，特殊边上的两个点称为特殊点，我们只需要比较特殊边的数量和特殊点被计算几次就可以了，如果特殊边和特殊点被计算的次数相同，则这就是我们需要找的点。

举个栗子，3，1 2，2 3，1 2 1这组数据，特殊边为1 2，2 3，数量为2，而1被计算了1次,2被计算了两次，3被计算了1次，所以2就是我们需要找的点，脑中模拟一下这个图就明白了。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int a[300000];int b[300000];int c[300000];int k[300000];void Init(){    memset(c,0,sizeof(c));}int main(){    int n;    while(~scanf("%d",&n)){        Init();        for(int i=0;i<n-1;i++){            scanf("%d%d",&a[i],&b[i]);        }        for(int i=1;i<=n;i++){            scanf("%d",&k[i]);        }        int ans=0;        for(int i=0;i<n-1;i++){            if(k[a[i]]!=k[b[i]]){                ans++;                c[a[i]]++;                c[b[i]]++;            }        }        bool flag=false;        for(int i=1;i<=n;i++){            if(c[i]==ans){                flag=true;                printf("YES\n");                printf("%d\n",i);                break;            }        }        if(!flag){            cout<<"NO"<<endl;        }    }    return 0;}