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 <bits/stdc++.h>
using namespace std;
#define F(i, s, t) for (int i = (s), _ = (t); i <= _; i ++)
#define Fo(i, s, t) for (int i = (s), _ = (t); i < _; i ++)
#define LL long long
const int N = 20;
const int S = 40000;
const int P = 1000000007;
int n, m, gr[N];
int p2[N*N], fu, cnt[S], edc[N][S];
int f[S];
int rd() {
	int x = 0; char c;
	do { c = getchar(); } while (!isdigit(c));
	do { x = x*10+c-'0', c = getchar(); } while (isdigit(c));
	return x;
}
int Ju(int s) { return (s & 3) == 0 || (s & 3) == 3; }
int main() {
	n = rd(), m = rd();
	F(i, 1, m) {
		int x = rd(), y = rd();
		gr[x] |= (1 << y-1);
	}
	p2[0] = 1;
	F(i, 1, n*n) p2[i] = p2[i-1] * 2 % P;
	fu = 1 << n;
	Fo(i, 1, fu) cnt[i] = cnt[i - (i & -i)] + 1;
	F(i, 1, n)
		Fo(s, 0, fu)
			edc[i][s] = cnt[gr[i] & s];
	f[0] = 1;
	Fo(s, 1, fu) if (Ju(s)) {
		for (int a = (s-1) & s; ; a = (a-1) & s) if (Ju(a) && f[a]) {
			int b = s-a, m = 1;
			F(i, 1, n)
				if (a >> i-1 & 1)
					m = (LL)m * (p2[edc[i][b]] - 1) % P;
			F(i, 1, n)
				if (b >> i-1 & 1)
					m = (LL)m * p2[edc[i][a]] % P;
			f[s] = (f[s] + (LL)f[a] * m) % P;
			if (!a) break;
		}
	}
	cout << (p2[m] - f[fu-1] + P) % P << endl;
	return 0;
}