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>
#define bi(x) (1<<x-1)
#define ll long long
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=15,mod=1e9+7;
int f[1<<N],e[N+5],w[N+2][1<<N],g[N+2][1<<N];
int n,m,i,j,k,x,p,u,v,sn,s,ans,ok[1<<N];
int main()
{
//	freopen("dag.in","r",stdin);
//	freopen("dag.out","w",stdout);
	scanf("%d%d",&n,&m);
	rep(i,1,m)
	{
		scanf("%d%d",&u,&v);
		e[u]|=bi(v);
	}
	sn=(1<<n)-1;
	rep(i,1,n)
	{
		w[i][0]=g[i][0]=1;
		rep(j,1,sn)
		{
			p=j&-j;
			if (!(e[i]&p)) w[i][j]=w[i][j^p];
			else w[i][j]=w[i][j^p]<<1;
			g[i][j]=w[i][j]-1;
		}
	}
	f[0]=1;
	rep(i,0,sn) if (!(i&1)==!(i&2)) ok[i]=1;
	rep(i,1,sn)
	if (ok[i])
	{
		for (j=i;j;j=j-1&i)
			if (ok[j])
			{
				k=i^j;
				s=f[k];
				rep(x,1,n)
					if (k&bi(x))
						s=(ll)s*g[x][j]%mod;
				rep(x,1,n)
					if (j&bi(x))
						s=(ll)s*w[x][k]%mod;
				f[i]=(f[i]+s)%mod;
			}
	}
	ans=1;
	rep(i,1,m) ans=(ll)ans*2%mod;
	printf("%d\n",(ans+mod-f[sn])%mod);
	return 0;
}