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>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)
#define x first
#define y second

using namespace std;

typedef long long LL;

const int N = 16;
const int Mod = 1e9 + 7;

inline void add(int &x, int y){
	x += y;
	if(x >= Mod) x -= Mod;
}

int mul(LL x, int y){
	return x * y % Mod;
}

int n, m;
int dp[1 << N], Pow[N * N];
int cnt[N][1 << N];

int main(){
	scanf("%d%d", &n, &m);

	For(i, 1, m){
		int u, v;
		scanf("%d%d", &u, &v);
		For(j, 0, (1 << n) - 1) if(j & (1 << (v - 1)))
			cnt[u][j]++;
	}

	Pow[0] = 1;
	For(i, 1, n * n) Pow[i] = mul(Pow[i - 1], 2);

	int ans = 0;
	For(i, 0, (1 << n) - 1) if(((i & 3) == 3) || !(i & 3)){
		dp[i] = 1;
		for(int j = (i - 1) & i; j; j = (j - 1) & i) if((j & 3) == 3 || !(j & 3)){
			int ret = dp[i ^ j];
			For(k, 0, n - 1) 
				if((i ^ j) & (1 << k)) ret = mul(ret, Pow[cnt[k + 1][j]] - 1);
				else if(j & (1 << k)) ret = mul(ret, Pow[cnt[k + 1][i ^ j]]);
			add(dp[i], ret);
		}
	}

	printf("%d\n", (Pow[m] - dp[(1 << n) - 1] + Mod) % Mod);
	return 0;
}