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;
int dp[(1<<15)+3],must=3;
int n,m;
int g[20][20],lg[(1<<15)+3];
int num[20][(1<<15)+3];
int pw[200];
const int mod=1e9+7;
int main(){
    //read
    scanf("%d%d",&n,&m);
    for(int i=1,f,t;i<=m;i++)
        scanf("%d%d",&f,&t),g[f-1][t-1]=1;
    //pre
    for(int i=0;i<n;i++) lg[1<<i]=i;
    pw[0]=1;for(int i=1;i<200;i++) pw[i]=pw[i-1]*2%mod;
    for(int i=0;i<n;i++)
        for(int S=1,t;S<(1<<n);S++)
            t=(S)&(-S),
            num[i][S]=num[i][S^t]+g[i][lg[t]];
    //solve
    dp[0]=1;
    for(int i=0;i<(1<<n);i++)
        for(int mask=(1<<n)-1-i,j=mask;j;j=(j-1)&mask)
            if( ( 1^(j&1) ^ ( (j&2) >>1) ) && dp[i]!=0 ){
                long long ways=1;
                for(int k=0;k<n;k++)
                    if(i&(1<<k)) (ways*=pw[num[k][j]]-1)%=mod;
                    else if(j&(1<<k)) (ways<<=num[k][i])%=mod;
                dp[i|j]=(dp[i|j]+dp[i]*ways)%mod;
            }
    
    cout<<((pw[m]-dp[(1<<n)-1])%mod +mod)%mod <<endl;
    return 0;
}