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 n,m;
typedef long long ll;
const int mod=1e9+7;
inline int add(int a,int b){a+=b;return a>=mod?a-mod:a;}
inline int sub(int a,int b){a-=b;return a<0?a+mod:a;}
inline int mul(int a,int b){return (ll)a*b%mod;}
int e[20];
int f[1<<16];

int main()
{
	cin >> n >> m;
	for(int i=1,u,v;i<=m;i++){
		scanf("%d%d",&u,&v);--u,--v;
		e[u]|=1<<v;
	}
	f[0]=1;
	for(int i=3;i<1<<n;i++)if((i&1)==((i&2)>>1)){
		for(int j=i;j;j=(j-1)&i)if((j&1)==((j&2)>>1)){
			int T0=j,T1=i^j,w=1;
			for(int id=0;id<n;id++)if(1&(i>>id)){
				if(1&(T0>>id)) w=mul(w,1<<(__builtin_popcount(e[id]&T1)));
				else w=mul(w,(1<<(__builtin_popcount(e[id]&T0)))-1);
			}
			f[i]=add(f[i], mul(f[T1],w));
		}
	}
	int tot=1;
	for(int i=1;i<=m;i++)tot=mul(tot,2);
	cout << sub(tot,f[(1<<n)-1]) << endl;
}
