764C Timofey and a tree[思维]

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

题意：给出n个点，n-1条边的有向树。并依次给出每个点所对应的颜色，问能否去掉一个掉，使得胜于的分支各自颜色相同。

思路：n的范围1e5 若枚举所有点显然不可行。观察得，如果保证剩余的分支颜色各自相同，那么如果一条边两边的点颜色不同，那么必定又一点被去掉。

遍历所有的边，如果两侧颜色不同，则记录两侧点，如果所有的这样的边都连接在同一个点上，则便是这个点，若不是，则没有答案。

代码：

#include<iostream>#include<algorithm>#include<cstring>#include<map>#include<set>#include<vector>using namespace std;int col[100005];vector<int> G[100005];map<int, int> s;int n;int main(){    cin >> n;    for(int i = 1; i <= n-1; ++i)    {        int u, v;        cin >> u >> v;        G[u].push_back(v);    }    for(int i = 1; i <=n; ++i)        cin >> col[i];    int cnt = 0;    vector<int>::iterator p;    for(int i = 1; i <= n; ++i)        for(p = G[i].begin(); p != G[i].end(); ++p)        {            if(col[i] != col[*p])            {                s[i]++;                s[*p]++;                cnt++;            }        }    for(int i = 1; i <= n; ++i)    {        if(s[i] == cnt)        {            cout << "YES" << endl;            cout << i;            return 0;        }    }    cout << "NO";    return 0;}