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 <iostream>
#include <vector>
#define llint long long
#define mod 1000000007

using namespace std;

llint n, m;
vector<int> G[20];
llint num[20][1<<15];
llint dp[1<<15];

int main(void)
{
	cin >> n >> m;
	llint u, v;
	for(int i = 0; i < m; i++){
		cin >> u >> v;
		u--, v--;
		G[u].push_back(v);
	}
	
	int S = 1<<n;
	for(int i = 0; i < n; i++){
		for(int j = 0; j < S; j++){
			for(int k = 0; k < G[i].size(); k++){
				if(j & (1<<G[i][k])) num[i][j]++;
			}
		}
	}
	
	dp[0] = 1;
	for(int i = 0; i < S; i++){
		int T = (S-1)-i;
		for(int j = T; j >= 1; j=(j-1)&T){
			if((j&3) == 3) continue;
			llint tmp = dp[i];
			for(int k = 0; k < n; k++){
				if(j & (1<<k)) continue;
				if(i & (1<<k)){
					tmp *= (1<<num[k][j]), tmp %= mod;
					continue;
				}
				tmp *= (1<<num[k][j]) - 1, tmp %= mod;
			}
			dp[i|j] += tmp, dp[i|j] %= mod;
		}
	}
	cout << dp[S-1] << endl;
	
	return 0;
}