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>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
const int mod=1e9+7;
int n,m,g[15],dp[1<<15],pw[235];
int main()
{
	int i,j,k;
	pw[0]=1;rep(i,230) pw[i]=pw[i-1]*2ll%mod;
	scanf("%d%d",&n,&m);rep(i,m){int x,y;scanf("%d%d",&x,&y);x--;y--;g[x]|=(1<<y);}
	dp[0]=1;
	for(i=1;i<(1<<n);i++)
	{
		if((i&1)==((i>>1)&1))
		{
			for(j=i;j;j=((j-1)&i))
			{
				if((j&1)==((j>>1)&1))
				{
					int coef=1;
					for(k=0;k<n;k++) if((i>>k)&1)
					{
						if((j>>k)&1) coef=1ll*coef*(1<<__builtin_popcount(g[k]&(i^j)))%mod;
						else coef=1ll*coef*((1<<__builtin_popcount(g[k]&j))-1)%mod;
					}
					dp[i]=(dp[i]+1ll*dp[i^j]*coef)%mod;
				}
			}
		}
	}
	cout<<(pw[m]-dp[(1<<n)-1]+mod)%mod<<endl;
	return 0;
}