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>

const int MAXN = 15;
const int mod = 1000000007;
typedef long long LL;
void reduce(int & x) { x += x >> 31 & mod; }
int out[MAXN], n, m;
int f[1 << 15];
inline bool g(int x) { return (x & 3) == 0 || (x & 3) == 3; }
inline int h(int x) { return 1 << __builtin_popcount(x); }
#define X (T >> j & 1)
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m;
	for (int i = 1, x, y; i <= m; ++i)
		std::cin >> x >> y, out[--x] |= 1 << --y;
	f[0] = 1; const int U = 1 << n;
	int ans = 1; while (m--) reduce(ans += ans - mod);
	for (int i = 1; i != U; ++i) if (g(i)) {
		for (int S = i; S; S = S - 1 & i) if (g(S)) {
			int T = i - S, t = 1;
			for (int j = 0; j != n; ++j) if (i >> j & 1)
				t = (LL) t * (h(out[j] & (X ? S : T)) - X) % mod;
			reduce(f[i] += (LL) t * f[T] % mod - mod);
		}
	}
	reduce(ans -= f[U - 1]);
	std::cout << ans << '\n';
	return 0;
}
