Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=15;
const int mod=1e9+7;
int n,m,f[1<<N],cnt[N+5][1<<N],g[N+5][1<<N],ans=1;
bool st[N+5][1<<N],ok[1<<N];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		st[u][1<<(v-1)]=true;
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<(1<<n);j++){
            int k=j&-j;
			cnt[i][j]=cnt[i][j^k]+st[i][k];
			g[i][j]=(1<<cnt[i][j])-1;
		}
	for(int i=0;i<(1<<n);i++)
        ok[i]=((i&1)>0)==((i&2)>0);
	f[0]=1;
	for(int i=1;i<(1<<n);i++)
		if(ok[i]){
			f[i]=1;
			for(int j=(i-1)&i;j;j=(j-1)&i)
				if(ok[j]){
					int k=i^j,s=f[k];
					for(int x=0;x<n;x++)
						if((1<<x)&k)
                            s=1ll*s*g[x+1][j]%mod;
						else if((1<<x)&j)
                            s=1ll*s*(1<<cnt[x+1][k])%mod;
					f[i]=(f[i]+s)%mod;
				}
		}
	for(int i=1;i<=m;i++)
        ans=ans*2%mod;
	ans=(ans-f[(1<<n)-1]+mod)%mod;
	printf("%d\n",ans);
}