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=18;
const int M=N*N;
const int mod=1e9+7;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,m,E[N],f[1<<N],pw[M],sz[1<<N];
int main() {
    n=gi(),m=gi();
    for(int i=1,u,v;i<=m;i++) u=gi()-1,v=gi()-1,E[u]|=1<<v;
    f[0]=1;
    for(int i=1;i<(1<<n);i++) sz[i]=sz[i>>1]+(i&1);
    pw[0]=1;for(int i=1;i<=m;i++) pw[i]=2ll*pw[i-1]%mod;
    for(int s=1;s<(1<<n);s++)
        if((s&1)==(s>>1&1))
            for(int t=s;t;t=(t-1)&s)
                if((t&1)==(t>>1&1)) {
                    int cnt=f[s^t];
                    for(int i=0;i<n;i++)
                        if(s>>i&1) {
                            if(t>>i&1) cnt=1ll*cnt*pw[sz[E[i]&(s^t)]]%mod;
                            else cnt=1ll*cnt*(pw[sz[E[i]&t]]-1+mod)%mod;
                        }
                    f[s]=(f[s]+cnt)%mod;
                }
    printf("%d\n",(pw[m]-f[(1<<n)-1]+mod)%mod);
    return 0;
}
