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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
const int N=1000;
const int M=1e6;
const int mo=1e9+7;
int n,m,d[N+10],f[M+10],bitcnt[M+10];
struct edges {
    int u,v;
}e[N+10];
int Sum(int x,int y) {
    x+=y;
    return (x>=mo)?x-mo:x;
}
int Mul(int x,int y) {
    return (long long)x*y%mo;
}
int main() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++) {
        int u,v;
        scanf("%d%d",&u,&v);
        d[u]|=(1<<(v-1));
    }
    for (int i=0;i<(1<<n);i++) {
        bitcnt[i]=0;
        for (int j=0;j<n;j++)
            if (i&(1<<j)) bitcnt[i]++;
    }
    f[0]=1;
    for (int i=0;i<(1<<n)-1;i++) {
        int nx=((1<<n)-1)^i;
        for (int j=nx;j>0;j=(j-1)&nx) {
            if ((j&1)&&(j&2)) continue;
            int va=1;
            for (int k=1;k<=n;k++)
                if (!(j&(1<<(k-1)))&&!(i&(1<<(k-1)))) {
                    int cnt=bitcnt[d[k]&j];
                    va=Mul(va,(1<<cnt)-1);
                }
                else if (j&(1<<(k-1))) {
                    int cnt=bitcnt[d[k]&(((1<<n)-1)^i^j)];
                    va=Mul(va,(1<<cnt));
                }
            f[i|j]=Sum(f[i|j],Mul(f[i],va));
        }
    }
    printf("%d\n",f[(1<<n)-1]);
    return 0;
}
