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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <bitset>

using namespace std;
typedef long long LL;
const int maxn = 20;
const int mod = 1e9+7;

int c[maxn][1<<15],_[maxn][1<<15];
int n,m,dp[1<<15],S;

int fpm(LL p,int k) {
	LL res=1ll;
	while (k) {
		if (k&1) (res*=p)%=mod;
		(p*=p)%=mod;k>>=1;
	}
	return res;
}
int main()
{
	#ifdef Amberframe
		freopen("agc016f.in","r",stdin);
		freopen("agc016f.out","w",stdout);
	#endif
	scanf("%d %d",&n,&m);S=(1<<n)-1;
	for (int u,v,i=1;i<=m;i++)
		scanf("%d %d",&u,&v),u--,v--,
		c[u][1<<v]++,_[v][1<<u]++;
	for (int i=0;i<n;i++)
	for (int j=1;j<=S;j++) {
		int o=j&-j;
		c[i][j]=c[i][j^o]+c[i][o];
		_[i][j]=_[i][j^o]+_[i][o];
	}
	
	for (int i=1;i<=S;i++)
	if ((i&1)==(i>>1&1)) {
		int res=1;
		for (int j=i-1&i;j;j=j-1&i)
		if (int w=dp[j]) {
			for (int k=0;k<n;k++)
			if (j>>k&1) {
				w=(LL)w*(1ll<<_[k][j^i])%mod;
				w=(LL)w*((1ll<<c[k][j^i])-1)%mod;
			}
			res=(res+w)%mod;
		}
		dp[i]=res;
	}
	int ans=(fpm(2,m)-dp[S])%mod;
	printf("%d",ans<0?ans+mod:ans);
	return 0;
}