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
//2018-1-17
//miaomiao
//
#include <bits/stdc++.h>
using namespace std;

#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)

#define N 16
const int P = 1e9 + 7;

inline int Mod(int a){
	if(a < 0) a += P;
	return a >= P? a-P: a;
}
inline int Mul(int a, int b){return 1ll * a * b % P;}

inline int Calc(int a){
	return __builtin_popcount(a);
}

int f[1<<N], reach[N];

bool Check(int a){
	int u = a & 1, v = (a & 2) >> 1;
	return u == v;
}

int main(){
	int n, m, u, v, t, tf;

	scanf("%d%d", &n, &m);
	For(i, 1, m){
		scanf("%d%d", &u, &v);
		reach[u] |= 1 << (v-1);
	}

	f[0] = 1;
	For(i, 1, (1<<n)-1) if(Check(i)) for(u = i; u; u = (u-1)&i){
		if(!Check(u)) continue;

		t = i ^ u; tf = f[t];
		For(j, 0, n-1){
			if(u&(1<<j)) tf = Mul(tf, 1<<Calc(reach[j+1]&t));
			if(t&(1<<j)) tf = Mul(tf, (1<<Calc(reach[j+1]&u))-1);
		}
		
		f[i] = Mod(f[i] + tf);
	}

	int ans = 1;
	while(m--) ans = Mul(ans, 2);
	ans = Mod(ans - f[(1<<n)-1]);

	printf("%d\n", ans);
	return 0;
}