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
/*ID:PROG:LANG:*/
#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define MAXN 16
#define MAXM (1<<15)
using namespace std;
LL n,m,dp[MAXM],cnt[MAXN][MAXM];
bool mp[MAXN][MAXM],ok[MAXM];
LL rd(){
	LL x=0,tp=1;char c=getchar();
	while(c<'0' || c>'9'){if(c=='-')tp=-1;c=getchar();}
	while(c>='0' && c<='9'){x=(x*10+c-'0');c=getchar();}
	return x*tp;
}
LL quick_mod(LL a,LL x){LL res=1;while(x){if(x&1)(res*=a)%=MOD;(a*=a)%=MOD,x>>=1;}return res;}
int main(){
	// freopen("testdata.in","r",stdin);
	// freopen("1.out","w",stdout);
	int x,y;
	n=rd(),m=rd();
	int S=1<<n;dp[0]=1;
	for(int i=1;i<=m;i++){
		x=rd(),y=rd();
		mp[x][1<<(y-1)]=1;
	}
	for(int i=1;i<=n;i++){
		for(int sta=1;sta<S;sta++){
			int j=sta&-sta;
			cnt[i][sta]=cnt[i][sta^j]+mp[i][j];
		}
	}
	for(int sta=0;sta<S;sta++)ok[sta]=(((1&sta)>0)==((2&sta)>0));
	for(int sta=1;sta<S;sta++){
		if(ok[sta]){
			dp[sta]=1;
			for(int i=(sta-1)&sta;i;i=(i-1)&sta){//枚举子集
				if(ok[i]){
					LL j=sta^i,tmp=dp[j];
					for(int k=0;k<n;k++){
						if((j>>k)&1)(tmp*=(1<<cnt[k+1][i])-1)%=MOD;
						else if((i>>k)&1)(tmp*=(1<<cnt[k+1][j]))%=MOD;
					}
					(dp[sta]+=tmp)%=MOD;
				}
			}
		}
	}
	printf("%d\n",(quick_mod(2,m)-dp[S-1]+MOD)%MOD);
	// printf("%.2lf\n",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}