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*<sub class="lower-index">*i*</sub>.

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.
The first line contains single integer *n* (2 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — 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 *c*<sub class="lower-index">1</sub>, *c*<sub class="lower-index">2</sub>, ..., *c*<sub class="lower-index">*n*</sub> (1 ≤ *c*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>), denoting the colors of the vertices.
none
4
1 2
2 3
3 4
1 2 1 1
YES
2
3
1 2
2 3
1 2 3
YES
2
4
1 2
2 3
3 4
1 2 1 2
NO
3
2 1
2 3
1 2 3
YES
2
4
1 2
2 4
4 3
1 1 3 2
YES
4
2
1 2
1 1
YES
1
10
5 7
4 5
10 2
3 6
1 2
3 4
8 5
4 9
2 3
15 15 15 15 5 15 26 18 15 15
YES
5
8
1 2
1 3
3 5
3 6
1 4
4 7
4 8
1 3 1 1 1 1 1 2
NO
3
2 1
2 3
4 4 4
YES
1
3
1 2
1 3
1 2 2
YES
1
4
1 4
2 4
3 4
1 2 3 1
YES
4
4
1 2
1 3
1 4
1 2 3 4
YES
1
9
1 2
2 3
3 4
4 5
2 7
7 6
2 8
8 9
1 1 2 2 2 3 3 4 4
YES
2
3
2 1
2 3
4 4 5
YES
2
4
1 2
2 3
3 4
1 2 2 1
NO
// #include <atcoder/fenwicktree>
#include <bits/stdc++.h>

using namespace std;
// using namespace atcoder;

constexpr long long INF_LL = 2000000000000000000LL;
constexpr int INF = 100000000;
constexpr long long MOD = 998244353;

#define all(x) x.begin(), x.end()
#define REP(i, a, b) for(int i = a; i < b; i++)
#define rep(i, n) REP(i, 0, n)

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<P> vp;
typedef vector<ll> vl;

int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
int sign[2] = {1, -1};

template <class T> bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}

ll modpow(ll a, ll b, ll m) {
    if(b == 0)
        return 1;
    ll t = modpow(a, b / 2, m);
    if(b & 1) {
        return (t * t % m) * a % m;
    } else {
        return t * t % m;
    }
}

struct edge {
    int to;
    ll cost;
    edge(int t, ll c) { to = t, cost = c; }
};

typedef vector<vector<int>> graph;

// using mint = modint998244353;
// constexpr int MAX_COM = 100001;
// mint fac[MAX_COM], ifac[MAX_COM];
// void initfac() {
//     fac[0] = ifac[0] = 1;
//     REP(i, 1, MAX_COM) fac[i] = i * fac[i - 1];
//     REP(i, 1, MAX_COM) ifac[i] = 1 / fac[i];
// }
// mint nCr(int n, int r){
//     if(r < 0 || n < r) return 0;
//     return fac[n] * ifac[n - r] * ifac[r];
// }

// typedef int S;
// S op(S x, S y){ return max(x, y); }
// S e(){ return 0; }
// typedef int F;
// S mapping(F f, S x){ return f + x; }
// F composition(F f, F g){ return f + g; }
// F id(){ return 0; }

void solve(){
    int n;
    cin >> n;
    graph g(n);
    rep(i, n - 1){
        int a, b;
        cin >> a >> b;
        g[a - 1].push_back(b - 1);
        g[b - 1].push_back(a - 1);
    }
    vi c(n), d;
    rep(i, n) cin >> c[i];
    auto dfs = [&](auto dfs, int v) -> void{
        d[v] = c[v];
        for(int x : g[v]){
            if(d[x] == -1) {
                dfs(dfs, x);
                if(d[v] != d[x]){
                    d[v] = -2;
                }
            }
        }
    };
    rep(i, n){
        for(int x : g[i]){
            if(c[i] != c[x]){
                bool flg;
                d = vi(n, -1);
                dfs(dfs, i);
                flg = true;
                for(int y : g[i]){
                    if(d[y] == -2){
                        flg = false;
                        break;
                    }
                }
                if(flg){
                    cout << "YES" << endl;
                    cout << i + 1 << endl;
                    return;
                }
                d = vi(n, -1);
                dfs(dfs, x);
                flg = true;
                for(int y : g[x]){
                    if(d[y] == -2){
                        flg = false;
                        break;
                    }
                }
                if(flg){
                    cout << "YES" << endl;
                    cout << x + 1 << endl;
                    return;
                }
                cout << "NO" << endl;
                return;
            }
        }
    }
}

int main(){
    cin.tie(0);
    //std::ios_base::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(16);

    // initfac();
    int t;
    t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
}
