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 long long ll;
const int maxn=18;
const int maxm=150;
const int mod=1e9+7;
int a[maxn][maxn];
int dp[(1<<16)+10];
int n, m;

bool check(int x){ return  ((x&3)==3) || ((x&3)==0) ; }

void add(int& x,int y){ x+=y; if(x>=mod) x-=mod; }

int out[maxn];
ll pw[maxm];

void init(){
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(a[i][j]) out[i] |= (1<<(j-1));
	pw[0]=1;
	for(int i=1;i<=m;i++) pw[i]=pw[i-1]*2%mod;
}

int main(){

	scanf("%d%d", &n, &m);
	int x, y;
	for(int i=1;i<=m;i++){
		scanf("%d%d", &x, &y);
		a[x][y]=1;
	}
	init();
	dp[0]=1;
	for(int i=1;i<(1<<n);i++)if(check(i)){
		for(int s=i;s;s=(s-1)&i)if(check(s)){
			int t=i^s, cnt=0, c=1;
			for(int j=1;j<=n;j++) if(s & (1<<(j-1))) cnt+=__builtin_popcount(out[j]&t);
			for(int j=1;j<=n;j++) if(t & (1<<(j-1))) c=c*(pw[ __builtin_popcount(out[j]&s) ]-1)%mod;
			add(dp[i], pw[cnt]*dp[t]%mod*c%mod);
		}
	}
	printf("%lld\n", (pw[m]-dp[(1<<n)-1]+mod)%mod);
	return 0;
}