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 <cstdio>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)
using namespace std;

typedef long long LL;
const int p=1000000007;
int n,m,f[33000],cnt[20][33000];
bool map[20][20];

int power(int x,int k)
{
    int ret=1;
    for (; k; k>>=1,x=LL(x)*x%p)
        if (k&1)
            ret=LL(ret)*x%p;
    return ret;
}

int main()
{
    scanf("%d%d",&n,&m);
    repu(i,1,m)
    {
        int u,v;
        scanf("%d%d",&u,&v),map[u-1][v-1]=1;
    }
    repu(i,0,n-1)
        repu(j,0,(1<<n)-1)
        {
            cnt[i][j]=1;
            repu(k,0,n-1)
                if ((j&(1<<k)) && map[i][k])
                    (cnt[i][j]*=2)%=p;
            --cnt[i][j];
            repu(k,0,n-1)
                if ((j&(1<<k)) && map[k][i])
                    (cnt[i][j]*=2)%=p;
        }
    f[0]=1;
    repu(i,0,(1<<n)-1)
        if (f[i])
        {
            int s=(1<<n)-1-i;
            for (int j=s; j; j=(j-1)&s)
                if ((j&1)==((j>>1)&1))
                {
                    int w=f[i];
                    repu(k,0,n-1)
                        if ((i>>k)&1)
                            w=LL(w)*cnt[k][j]%p;
                    (f[i^j]+=w)%=p;
                }
        }
    printf("%d\n",(power(2,m)-f[(1<<n)-1]+p)%p);
    return 0;
}