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<iostream>
#include<cstdio>
#include<cstring>

namespace ShangWuKeNengXieBuWanLe____XiaWu____no__ZhongWuJiXuXieBa
{
	typedef long long ll;
	const int N=15,M=N*N,MOD=1000000007;
	inline void inc(int a,int &b){b=(a+b)%MOD;}
	ll qpow(ll a,ll b){ll c=1;for(;b;b>>=1,a=a*a%MOD)if(b&1)c=c*a%MOD;return c;}

	int begin[N+2],next[M],to[M];
	int n,m,e,S;
	void add(int x,int y)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
	}
	void initialize()
	{
		scanf("%d%d",&n,&m);
		for(int i=1,u,v;i<=m;i++)
			scanf("%d%d",&u,&v),add(u,v);
		S=1<<n;
	}

	int f[1<<N];

	inline bool ins(int A,int p){return (A>>(p-1))&1;}
	int calc(int A,int B)
	{
		int ret=1;
		for(int p=1;p<=n;p++)
			if(ins(A,p))
			{
				int tmp=0;
				for(int i=begin[p],q;i;i=next[i])
					tmp+=ins(B,q=to[i]);
				tmp=qpow(2,tmp)-1;
				ret=(ll)ret*tmp%MOD;
			}
		for(int p=1;p<=n;p++)
			if(ins(B,p))
			{
				int tmp=0;
				for(int i=begin[p];i;i=next[i])
					tmp+=ins(A,to[i]);
				ret=(ll)ret*qpow(2,tmp)%MOD;
			}
		return ret;
	}

	inline bool precheck(int A){return !(ins(A,1)^ins(A,2));}
	void dp()
	{
		f[0]=1;
		for(int A=1;A<S;A++)
			for(int B=A;B;B=(B-1)&A)
				if(precheck(B))inc((ll)f[A^B]*calc(A^B,B)%MOD,f[A]);
	}
	void solve()
	{
		initialize();
		dp();
		int ans=qpow(2,m)-f[S-1];
		ans=(ans%MOD+MOD)%MOD;
		printf("%d\n",ans);
	}
}

int main()
{
	ShangWuKeNengXieBuWanLe____XiaWu____no__ZhongWuJiXuXieBa::solve();
	return 0;
}
