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>

using namespace std;

const int maxn = 15;
const int mod = 1e9+7;

int n, m;
int out[maxn], in[maxn], sz[1<<maxn];
int dp[1<<maxn];

int lowbit(int x) {return x & (-x);}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i < (1<<n); i++) sz[i] = sz[i ^ lowbit(i)] + 1;
	for (int i = 1; i <= m; i++) {
		int x, y; scanf("%d%d", &x, &y);
		-- x, -- y;
		out[x] ^= (1<<y), in[y] ^= (1<<x);
	}
	dp[0] = 1;
	for (int mask = 1; mask < (1<<n); ++ mask) {
		for (int s = mask; s; s = (s - 1) & mask) {
			if ((s & 1) && (s & 2)) continue;
			int t = dp[mask ^ s];
			for (int i = 0; i < n; i++) {
				if (s & (1<<i)) {
					t = 1LL * t * (1<<(sz[out[i] & (~ mask)])) % mod;
				} else if (!(mask & (1<<i))) {
					t = 1LL * t * ((1<<sz[s & out[i]]) - 1) % mod;
				}
			}
			dp[mask] = (dp[mask] + t) % mod;
		}
	}
	printf("%d\n", dp[(1<<n)-1]);
	return 0;
}