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 mod 1000000007
#define popcnt __builtin_popcount
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int edge[20],p[233],dp[1<<15|1];
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),m=gi(),u,v;
	for(int i=1;i<=m;++i)u=gi(),v=gi(),edge[u]|=1<<v-1;
	p[0]=1;for(int i=1;i<233;++i)p[i]=p[i-1]*2%mod;
	for(int i=1;i<1<<n;++i){
		if(((i>>1)&1)^(i&1))continue;
		dp[i]=1;
		for(int N=i,P;N;N=(N-1)&i){
			P=i^N;int mul=1;
			for(int j=1;j<=n;++j)if((N>>j-1)&1)mul=1ll*mul*p[popcnt(edge[j]&P)]%mod;
			for(int j=1;j<=n;++j)if((P>>j-1)&1)mul=1ll*mul*(p[popcnt(edge[j]&N)]-1)%mod;
			dp[i]=(dp[i]+1ll*dp[P]*mul)%mod;
		}
	}
	int ans=1;for(int i=1;i<=m;++i)ans=ans*2%mod;
	ans=(ans-dp[(1<<n)-1]+mod)%mod;
	printf("%d\n",ans);
	return 0;
}
