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 <bits/stdc++.h>
using namespace std;
const int N=16,M=N*N;
const int Mod=1000000007;
int G[N],Pow[M],f[1<<N],cnt[1<<N];
inline int read() {
    int x=0;
    char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
    return x;
}
int main() {
    int n=read(),m=read(),lim=1<<n; Pow[0]=f[0]=1;
    for (int i=1;i<=m;i++) Pow[i]=(Pow[i-1]<<1)%Mod;
    for (int i=0;i<lim;i++) cnt[i]=cnt[i>>1]+(i&1);
    for (int i=1;i<=m;i++) {int x=read(),y=read();G[x-1]|=1<<(y-1);}
    for (int sta=0;sta<lim;sta++) if ((sta&1)==(sta>>1&1))
        for (int T=sta;T;T=sta&(T-1)) if ((T&1)==(T>>1&1)) {
            int res=1,U=sta^T;
            for (int i=0;i<n;i++) if (sta>>i&1) {
                res=1LL*res*(T>>i&1?1<<cnt[G[i]&U]:(1<<cnt[G[i]&T])-1)%Mod;
            }
            (f[sta]+=1LL*f[U]*res%Mod)%=Mod;
        }
    printf("%d\n",(Pow[m]-f[lim-1]+Mod)%Mod);
    return 0;
}