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<cstdio>
using namespace std;
typedef long long ll;
const int N=22,M=1<<15,P=1000000007;
int i,j,k,n,m,x,y,sum;
int a[N],f[M],cnt[M];
int main() {
	scanf("%d%d",&n,&m);
	sum=1;
	for (i=0;i<m;i++) {
		sum=(sum+sum)%P;
		scanf("%d%d",&x,&y);
		x--;y--;
		a[x]|=1<<y;
	}
	m=1<<n;
	for (i=0;i<m;i++)
		for (j=0;j<n;j++) if (i&1<<j) cnt[i]++;
	f[0]=1;
	for (i=0;i<m;i++) if (((i&2)>>1)==(i&1)) {
		for (j=i;j;j=(j-1)&i) {
			int A=j,B=i^j;
			int ans=f[B];
			for (k=0;k<n;k++) {
				if (B&1<<k) ans=(ll) ans*((1<<cnt[a[k]&A])-1)%P;
				if (A&1<<k) ans=(ll) ans*(1<<cnt[a[k]&B])%P;
			}
			f[i]=(f[i]+ans)%P;
		}
	}
	printf("%d\n",(sum+P-f[m-1])%P);
}