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<cstdio>
#include<algorithm>

using namespace std;

const long long mod = 1000000007;

bool G[15][15];
int N;
int M;

void input(){
	scanf("%d%d", &N, &M);
	for(int i = 0; i < M; ++i){
		int x, y;
		scanf("%d%d", &x, &y);
		x--;
		y--;
		G[x][y] = true;
	}
}

long long dp[1 << 15];

inline long long f(int n){
	return (1 << n) - 1;
}

inline int count(int v, int S){
	int res = 0;
	for(int i = 0; i < N; ++i){
		if((S >> i) & 1){
			if(G[v][i]) res++;
		}
	}
	return res;
}

bool check(int S){
	int x = (S & 1) ^ ((S & 2) >> 1);
	return x == 0;
}

long long solve(){
	dp[0] = 1;
	int mask = (1 << N) - 1;
	for(int S = 1; S < (1 << N); ++S){
		if(!check(S)) continue;
		for(int S2 = (S - 1) & S; ; S2 = (S2 - 1) & S){
			if(!check(S2)) continue;
			long long coe = 1;
			for(int v = 0; v < N; ++v){
				if((S >> v) & 1) continue;
				int cnt = count(v, S - S2);
				coe *= f(cnt);
				coe %= mod;
			}
			for(int v = 0; v < N; ++v){
				int x = S - S2;
				if((x >> v) & 1){
					int cnt = count(v, mask - S);
					coe *= (f(cnt) + 1);
					coe %= mod;
				}
			}
			dp[S] += dp[S2] * coe;
			dp[S] %= mod;
			if(S2 == 0) break;
		}
	}
	long long all = 1;
	for(int i = 0; i < M; ++i){
		all *= 2;
		all %= mod;
	}
	long long ans = all - dp[mask];
	ans %= mod;
	ans += mod;
	ans %= mod;
	return ans;
}

int main(){
	input();
	long long ans = solve();
	printf("%lld\n", ans);
	return 0;
}
