We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and b_i. For each k=1, ..., N, solve the problem below:

* Consider writing a number on each vertex in the tree in the following manner:
* First, write 1 on Vertex k.
* Then, for each of the numbers 2, ..., N in this order, write the number on the vertex chosen as follows:
* Choose a vertex that still does not have a number written on it and is adjacent to a vertex with a number already written on it. If there are multiple such vertices, choose one of them at random.
* Find the number of ways in which we can write the numbers on the vertices, modulo (10^9+7).

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq a_i,b_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

For each k=1, 2, ..., N in this order, print a line containing the answer to the problem.

Examples

Input

3
1 2
1 3


Output

2
1
1


Input

2
1 2


Output

1
1


Input

5
1 2
2 3
3 4
3 5


Output

2
8
12
3
3


Input

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8


Output

40
280
840
120
120
504
72
72
#include <bits/stdc++.h>

using namespace std;

int n;
vector<int> adj[200010];
int sz[200010], dp[200010], ans[200010];
int f[200010], fi[200010];
const int MOD = 1e9 + 7;

int add() {return 0;}
template<typename... Args>
int add(long long a, Args... args){
    return (a + add(args...)) % MOD;
}

int mul() {return 1;}
template<typename... Args>
int mul(long long a, Args... args){
    return (a * mul(args...)) % MOD;
}


int fp(int a, int b){
    int ret = 1;

    while(b){
        if(b & 1) ret = mul(ret, a);

        a = mul(a, a);
        b >>= 1;
    }

    return ret;
}

int inv(int x){
    return fp(x, MOD - 2);
}

void gen(){
    f[0] = fi[0] = 1;

    for(int i = 1; i <= n; i++){
        f[i] = mul(i, f[i - 1]);
        fi[i] = inv(f[i]);
    }
}

void dfs(int u, int p){
    sz[u] = 1;
    dp[u] = 1;

    for(auto &v : adj[u]){
        if(p == v) continue;

        dfs(v, u);
        sz[u] += sz[v];

        dp[u] = mul(dp[u], dp[v], fi[sz[v]]);
    }

    dp[u] = mul(dp[u], f[sz[u] - 1]);
}

void dfs0(int u, int p, int dpPar){
    ans[u] = mul(dp[u], dpPar, fi[n - sz[u]], f[n - 1], fi[sz[u] - 1]);

    int val = mul(dp[u], dpPar, fi[n - sz[u]]);
    for(auto &v : adj[u]){
        if(v == p) continue;

        dfs0(v, u, mul(val, f[sz[v]], inv(dp[v]), f[n - sz[v] - 1], fi[sz[u] - 1]));
    }
}

int main(){
    scanf("%d", &n);

    gen();

    for(int i = 1; i < n; i++){
        int u, v;
        scanf("%d%d", &u, &v);

        adj[u].emplace_back(v);
        adj[v].emplace_back(u);
    }

    dfs(1, 1);
    dfs0(1, 1, 1);

    for(int i = 1; i <= n; i++){
        printf("%d\n", ans[i]);
    }
    
    return 0;
}