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
//Heaplax
#include<bits/stdc++.h>
#define bit(x) __builtin_popcount((x))
#define LL long long
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;
}
using namespace std;
const int mod=1000000007;
bool gg[1<<15];
int n,m,ans,f[1<<15],edge[15];
int main()
{
	re(n),re(m);
	ans=1;
	for(int i=1;i<=m;++i)
		ans=ans*2%mod;
	for(int i=1,u,v;i<=m;++i)
	{
		re(u),re(v);
		--u,--v;
		edge[u]|=1<<v;
	}
	for(int i=0;i<(1<<n);++i)
		if((i&1)^(i>>1&1))
			gg[i]=1;
	for(int i=0;i<(1<<n);++i)if(!gg[i])
	{
		f[i]=1;
		for(int j=(i-1)&i;j;j=(j-1)&i)if(!gg[j])
		{
			int k=i^j;
			int base=f[j];
			for(int x=0;x<n;++x)if(j>>x&1)
				base=1ll*base*((1<<bit(edge[x]&k))-1)%mod;
			for(int x=0;x<n;++x)if(k>>x&1)
				base=1ll*base*(1<<bit(edge[x]&j))%mod;
			f[i]=(f[i]+base)%mod;
		}
	}
	ans=(ans-f[(1<<n)-1]+mod)%mod;
	printf("%d\n",ans);
}
