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 <cstdio>
#include <cctype>
#include <algorithm>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 15, P = 1e9+7;
int n, m, f[1<<N|5], w[N*N], g[N], h[1<<N|5];
int main()
{
    n = read(), m = read(), f[0] = w[0] = 1;
    for(int i = 1; i <= m; ++i)
    {
        int x = read()-1, y = read()-1;
        g[x] |= 1<<y, w[i] = (w[i-1]<<1)%P;
    }
    for(int s = 1; s < 1<<n; ++s) h[s] = h[s^(s&-s)]+1;
    for(int s = 1; s < 1<<n; ++s)
        if((s&1) == (s>>1&1))
            for(int u = s; u; u = (u-1)&s)
                if((u&1) == (u>>1&1))
                {
                    int k = 1;
                    for(int i = 0; i < n; ++i)
                        if(s>>i&1)
                        {
                            if(u>>i&1) k = 1ll*k*w[h[g[i]&(s^u)]]%P;
                            else k = 1ll*k*(w[h[g[i]&u]]-1)%P;
                        }
                    f[s] = (f[s]+1ll*f[s^u]*k%P)%P;
                }
    printf("%d\n", (w[m]-f[(1<<n)-1]+P)%P);
    return 0;
}