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 MAXN	20
#define MAXS	65536
#define P	1000000007
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
vector <int> a[MAXN];
long long ans[MAXS];
long long f[MAXN][MAXS], g[MAXN][MAXS];
void update(long long &x, long long y) {
	x = (x + y) % P;
}
int main() {
	int n, m;
	read(n), read(m);
	for (int i = 1; i <= m; i++) {
		int x, y;
		read(x), read(y);
		a[x].push_back(y);
	}
	int goal = (1 << n) - 1;
	for (int s = 0; s <= goal; s++)
	for (int i = 1; i <= n; i++) {
		int tmp = 0;
		for (unsigned j = 0; j < a[i].size(); j++) {
			int tj = 1 << a[i][j] - 1;
			if (tj & s) tmp++;
		}
		f[i][s] = 1 << tmp;
		g[i][s] = f[i][s] - 1;
	}
	ans[0] = 1;
	for (int s = 1; s <= goal; s++) {
		if ((s & 1) != ((s & 2) >> 1)) continue;
		for (int t = s, b = 0; t != 0; t = (t - 1) & s, b = s ^ t) {
			if ((t & 1) != ((t & 2) >> 1)) continue;
			long long tmp = ans[b];
			for (int i = 1, j = 1; i <= n; i++, j <<= 1) {
				if (j & t) tmp = tmp * f[i][b] % P;
				if (j & b) tmp = tmp * g[i][t] % P;
			}
			update(ans[s], tmp);
		}
	}
	long long finalans = 1;
	for (int i = 1; i <= m; i++)
		finalans = finalans * 2 % P;
	finalans = (finalans - ans[goal] + P) % P;
	cout << finalans << endl;
	return 0;
}