Problem: There is a directed graph G with N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i is directed from x_i to y_i. Here, x_i < y_i holds. Also, there are no multiple edges in G.

Consider selecting a subset of the set of the M edges in G, and removing these edges from G to obtain another graph G'. There are 2^M different possible graphs as G'.

Alice and Bob play against each other in the following game played on G'. First, place two pieces on vertices 1 and 2, one on each. Then, starting from Alice, Alice and Bob alternately perform the following operation:

* Select an edge i such that there is a piece placed on vertex x_i, and move the piece to vertex y_i (if there are two pieces on vertex x_i, only move one). The two pieces are allowed to be placed on the same vertex.



The player loses when he/she becomes unable to perform the operation. We assume that both players play optimally.

Among the 2^M different possible graphs as G', how many lead to Alice's victory? Find the count modulo 10^9+7.

Constraints

* 2 ≤ N ≤ 15
* 1 ≤ M ≤ N(N-1)/2
* 1 ≤ x_i < y_i ≤ N
* All (x_i,\ y_i) are distinct.

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the number of G' that lead to Alice's victory, modulo 10^9+7.

Examples

Input

2 1
1 2


Output

1


Input

3 3
1 2
1 3
2 3


Output

6


Input

4 2
1 3
2 4


Output

2


Input

5 10
2 4
3 4
2 5
2 3
1 2
3 5
1 3
1 5
4 5
1 4


Output

816

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;

const int MOD = 1000000007;

int g[15];
int dp[1 << 15];

int main() {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = 0; i < m; i ++) {
                int a, b;
                scanf("%d%d", &a, &b);
                a --, b --;
                g[a] |= (1 << b);
        }               
        dp[0] = 1;
        for (int ST = 1; ST < (1 << n); ST ++) {
                int U = ((1 << n) - 1) ^ ST;
                for (int S = ST - 1; S >= 0; S --) {
                        S &= ST;
                        int T = ST ^ S;
                        if ((T & 3) == 3) continue;
                        long long res = 1;
                        for (int i = 0; i < n; i ++) {
                                if (T & (1 << i)) (res *= 1 << __builtin_popcount(g[i] & U)) %= MOD;
                                if (U & (1 << i)) (res *= (1 << __builtin_popcount(g[i] & T)) - 1) %= MOD;
                        }
                        (dp[ST] += res * dp[S] % MOD) %= MOD;
                }
        }
        printf("%d\n", dp[(1 << n) - 1]);
        return 0;
}