A. 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 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

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

题意:

一棵树中各个节点被染上了颜色;

让你在一棵树中随便选一个节点作为根节点,Tim不看这个根节点的颜色，但是这个根节点下的子树们不满足要求的话会惹怒他，问你是否存在一个根节点，这个根节点的直系儿子节点的子树里面的所有节点的颜色都一样; 能输出YES，第二行输出这个根的下标。否则输出NO。

注意不同的子树颜色可以不一样，但子树里面的点颜色要相同。

思路：

搜索，最短路，并查集都可以做吧。。。但是有个思路更简单！

在整张图;那些边的两端端点颜色不一样的边，设为特殊边,用计数器把特殊边的数目记录下来，这样的边肯定是有和根节点连在一起的;

因为如果没有和根节点相连的话,肯定会造成子树里面有两个颜色不一样的

所以就看看哪个节点和m条特殊边都相连,如果有的话肯定就是它作为根节点了,且如果没有这样的点的话肯定无解了

#include 
    
     using namespace std;#define LL long long#define mp make_pair#define pb push_backtypedef pair
     
       pii;typedef pair
      
        pll;const double pi = acos(-1.0);const int MAXN = 1e5+10;int n,m;int h[100010],color[MAXN];pii ee[MAXN];int main(){        scanf("%d",&n);    for(int i = 1; i <= n-1; i++)    {        scanf("%d",ee[i].first);        scanf("%d",ee[i].second);    }    for(int i = 1; i <= n; i++)    {        scanf("%d",color[i]);    }    for(int i = 1; i <= n-1; i++){        int x = ee[i].first,y = ee[i].second;        if (color[x]!=color[y])        {            m++;            h[x]++,h[y]++;        }    }    for(int i = 1; i <= n; i++)        if (h[i]==m)        {            puts("YES");s            return 0;        }    puts("NO");    return 0;}