C. Timofey and a tree

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 c_i.

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, ..., c_n (1 ≤ c_i ≤ 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

4
1 2
2 3
3 4
1 2 1 1

Output

YES
2

Input

3
1 2
2 3
1 2 3

Output

YES
2

Input

4
1 2
2 3
3 4
1 2 1 2

Output

NO

这么简单的问题，但是我还是没有想到，或者说是想复杂了，但是这个问题好像给了我的新的idea，好像就是一种独特的idea，我们思考问题的时候总是应该从它最核心的地方来
寻找的，或者说是作者想让我们干什么，我们发现这个题就是我们能改变的就是一条边连着两个不相同的点，这两个点我们必须要使一个变成根节点，要是我们不使这个点变成根节
点的话，我们是得不到符合要求的东西的。所以这个题有两种解决的办法，第一种就是找到一个边连接着权值不相同的两个点要是没有找到的话肯定是输出YES，然后的话我们把这两个
点每一个都dfs一下，要是可以的话就可以，否则的话就是输出NO。还有一个简单的做法就是就是画个图你就明显的看出来了。

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int ans[N],w[N];
struct node
{
    int a,b;
}s[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        scanf("%d%d",&s[i].a,&s[i].b);
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&w[i]);
    }
    int cnt=0;
    for(int i=1;i<n;i++)
    {
        int x=s[i].a; int y=s[i].b;
        if(w[x]==w[y]) continue;
        ans[x]++; ans[y]++;
        cnt++;
    }
    int flag=0;
    for(int i=1;i<=n;i++)
    {
        if(ans[i]==cnt)
        {
            flag=1;
            printf("YES
");
            printf("%d
",i); return 0;
        }
    }
    printf("NO
");
}