【codeforces 764C】Timofey and a tree

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

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

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

【题解】



题意:

一棵树中各个节点被染上了c[i]颜色;

让你在一棵树中随便选一个节点作为根节点,然后把整棵树抬起来;

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;

做法:

考虑最后整张图;

那些边的两端端点颜色不一样的边(设为特殊边,这样的边总数为m)肯定是有和根节点连在一起的;

(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)



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 1e5+10;

int n,m;
int h[100010],c[MAXN];
pii bian[MAXN];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n-1)
        rei(bian[i].fi),rei(bian[i].se);
    rep1(i,1,n)
        rei(c[i]);
    rep1(i,1,n-1)
    {
        int x = bian[i].fi,y = bian[i].se;
        if (c[x]!=c[y])
        {
            m++;
            h[x]++,h[y]++;
        }
    }
    rep1(i,1,n)
        if (h[i]==m)
        {
            puts("YES");
            printf("%d\n",i);
            return 0;
        }
    puts("NO");
    return 0;
}