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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const ll MOD = 1000000007;

int N, M;
ll p2[310];
bool e[16][16];
ll dp[1 << 15];

int main() {
	p2[0] = 1;
	for (int i = 1; i < 310; ++i) {
		p2[i] = p2[i-1] * 2 % MOD;
	}

	cin >> N >> M;

	rep(i, M) {
		int x, y;
		cin >> x >> y;
		--x; --y;
		e[x][y] = 1;
	}

	dp[0] = 1;

	for (int i = 1; i < (1 << N); ++i) {
		for (int j = i; j >= 0; --j) {
			j &= i;

			int nx = j ^ i;
			if (dp[j] == 0 || ((nx & 1) && (nx & 2))) {
				continue;
			}

			int rem = ((1 << N) - 1) ^ i;

			ll way = 1;

			rep(k, N) {
				if ((rem >> k) & 1) {
					int cnt = 0;
					rep(t, N) if ((nx >> t) & 1) {
						if (e[k][t]) {
							++cnt;
						}
					}
					way = way * (p2[cnt] - 1) % MOD;
				} else if ((j >> k) & 1) {
					int cnt = 0;
					rep(t, N) if ((nx >> t) & 1) {
						if (e[k][t]) {
							++cnt;
						}
					}
					way = way * p2[cnt] % MOD;	
				}
			}

			dp[i] = (dp[i] + dp[j] * way) % MOD;
		}
	}


	cout << dp[(1 << N) - 1] << endl;

    return 0;
}