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 rep(i, a, b) for (int i = (a); i <= int(b); i++)
#define per(i, a, b) for (int i = (a); i >= int(b); i--)
#define fir first
#define sec second
#define tct template<class type>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
const int maxn = 1 << 15, mod = 1e9 + 7;
int n, m, a[20], f[maxn + 5];

inline void red(int &x) { x += x >> 31 & mod; }
tct inline void cmax(type &x, type y) { x < y ? x = y : 0; }
tct inline void cmin(type &x, type y) { x > y ? x = y : 0; }

int main() {
	scanf("%d %d", &n, &m);
	int sum = 1;
	rep(i, 1, m) sum = ll(2) * sum % mod;
	for (int i = 0, x, y; i < m; i++) {
		scanf("%d %d", &x, &y);
		a[x] |= 1 << (y - 1);
	}
	f[0] = 1;
	rep(msk, 1, (1 << n) - 1) {
		for (int old = msk; old; old = (old - 1) & msk) {
			if ((old & 1) ^ (old >> 1 & 1)) continue;
			int pi = 1;
			rep(j, 1, n) {
				if ((msk ^ old) >> (j - 1) & 1) pi = ll((1 << __builtin_popcount(a[j] & old)) - 1) * pi % mod;
				else if (old >> (j - 1) & 1) pi = ll((1 << __builtin_popcount(a[j] & (msk ^ old)))) * pi % mod;
			}
			f[msk] = (f[msk] + ll(pi) * f[msk ^ old]) % mod;
		}
	}
	red(sum -= f[(1 << n) - 1]);
	printf("%d\n", sum);
	return 0;
}