Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 16
using namespace std;
const int mod = 1e9+7;
inline int read(){
	int sum = 0, t = 1; char ch = getchar();
	while(ch < '0' || ch > '9'){ if(ch == '-') t = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9'){ sum = sum * 10 + ch - '0'; ch = getchar(); }
	return sum * t;
}

int n, m, G[N];
int cnt[1<<N];
int f[1<<N];

void upd(int &x, int y){ x += y; if(x >= mod) x -= mod; }

int main(){
	n = read(); m = read();
	for(int S = 1; S < 1 << n; S++) cnt[S] = cnt[S ^ (S & -S)] + 1;
	for(int i = 1; i <= m; i++){
		int x = read() - 1, y = read() - 1;
		G[x] |= 1 << y;
	}
	f[0] = 1;
	for(int S = 0; S < 1 << N; S++) if((S >> 1 & 1) == (S & 1))
		for(int T = S; T ; T = T - 1 & S) if((T >> 1 & 1) == (T & 1)){
			int tmp = 1;
			for(int i = 0; i < n; i++) if(S >> i & 1){
				if(T >> i & 1) tmp = 1LL * tmp * (1 << cnt[G[i] & (S ^ T)]) % mod;
				else tmp = 1LL * tmp * ((1 << cnt[G[i] & T]) - 1) % mod;
			}
			upd(f[S], 1LL * f[S ^ T] * tmp % mod);
		}
	int pw = 1;
	for(int i = 1; i <= m; i++) pw = 1LL * pw * 2 % mod;
	printf("%d\n", (pw - f[(1 << n) - 1] + mod) % mod);
}