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>

typedef long long LL;

const int N=15,M=N*N,Mod=1e9+7;

int n,m,x[M],y[M],dp[1<<N],out[N],qp[M],cnt[1<<N];

void Init() {
	scanf("%d%d",&n,&m);
	for (int i=0;i<m;++i) scanf("%d%d",&x[i],&y[i]),x[i]--,y[i]--;
}

void Solve() {
	int ans=0;
	for (int i=0;i<m;++i) out[x[i]]|=1<<y[i];
	qp[0]=1; for (int i=1;i<=m;++i) qp[i]=qp[i-1]*2%Mod;
	int Max=1<<n; dp[0]=1;
	for (int i=0;i<Max;++i) cnt[i]=cnt[i>>1]+(i&1);
	for (int s=0;s<Max;++s) if ((s&3)==3 || (s&3)==0) {
		for (int w=s;w;w=(w-1)&s) if (dp[s^w] && ((w&3)==3 || (w&3)==0)) {
			LL sm=dp[s^w]; int t=s^w;
			for (int i=0;i<n;++i) {
				if ((1<<i)&w) sm=1ll*sm*qp[cnt[t&out[i]]]%Mod;
				else if ((1<<i)&t) sm=1ll*sm*(qp[cnt[w&out[i]]]-1)%Mod;
			}
			dp[s]=(dp[s]+sm)%Mod;
		}
		ans=(ans+dp[s])%Mod;
	} 
 	printf("%d\n",(qp[m]-dp[Max-1]+Mod)%Mod);
}

int main() {
	Init();
	Solve();
	return 0;
}