There is a tree with N vertices, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.

Taro has decided to paint each vertex in white or black, so that any black vertex can be reached from any other black vertex by passing through only black vertices.

You are given a positive integer M. For each v (1 \leq v \leq N), answer the following question:

* Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.

Constraints

* All values in input are integers.
* 1 \leq N \leq 10^5
* 2 \leq M \leq 10^9
* 1 \leq x_i, y_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}


Output

Print N lines. The v-th (1 \leq v \leq N) line should contain the answer to the following question:

* Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.

Examples

Input

3 100
1 2
2 3


Output

3
4
3


Input

4 100
1 2
1 3
1 4


Output

8
5
5
5


Input

1 100


Output

1


Input

10 2
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7


Output

0
0
1
1
1
0
1
0
1
1
def main():
    import sys
    input = sys.stdin.readline

    N, mod = map(int, input().split())
    adj = [[] for _ in range(N+1)]
    for _ in range(N-1):
        a, b = map(int, input().split())
        adj[a].append(b)
        adj[b].append(a)

    """ここから変更"""
    # op: 普通に根を固定した木DPで子の情報を親に集めるときの演算
    def op(a, b):
        return (a * (b + 1))%mod

    # cum_merge: 累積opどうしをマージするときの演算
    def cum_merge(a, b):
        return (a * b)%mod
    """ここまで変更"""

    # root=1でまず普通に木DPをする
    # 並行して各頂点につき、子の値の累積opを左右から求めておく
    # その後根から順番に、親からの寄与を求めていく(from_par)
    def Rerooting(adj):
        N = len(adj) - 1
        st = [1]
        seen = [0] * (N + 1)
        seen[1] = 1
        par = [0] * (N + 1)
        child = [[] for _ in range(N + 1)]
        seq = []
        while st:
            v = st.pop()
            seq.append(v)
            for u in adj[v]:
                if not seen[u]:
                    seen[u] = 1
                    par[u] = v
                    child[v].append(u)
                    st.append(u)
        seq.reverse()
        dp = [1] * (N + 1)
        left = [1] * (N + 1)
        right = [1] * (N + 1)
        for v in seq:
            tmp = 1
            for u in child[v]:
                left[u] = tmp
                tmp = op(tmp, dp[u])
            tmp = 1
            for u in reversed(child[v]):
                right[u] = tmp
                tmp = op(tmp, dp[u])
            dp[v] = tmp

        seq.reverse()
        from_par = [0] * (N + 1)
        for v in seq:
            if v == 1:
                continue
            from_par[v] = op(cum_merge(left[v], right[v]), from_par[par[v]])
            dp[v] = op(dp[v], from_par[v])

        return dp

    dp = Rerooting(adj)
    for i in range(1, N+1):
        print(dp[i])


if __name__ == '__main__':
    main()
