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>
#define MAXN 15
#define MAXS 50000
#define MO 1000000007
using namespace std;
int n,m,all;
int cnt[MAXN+5][MAXS+5],p[MAXN+5][MAXS+5];
int dp[MAXS+5];
bool legal[MAXS+5];
int PowMod(int a,int b)
{
	int ret=1;
	while(b)
	{
		if(b&1)
			ret=1LL*ret*a%MO;
		a=1LL*a*a%MO;
		b>>=1;
	}
	return ret;
}
int LowBit(int i)
{
	return i&(-i);
}
int main()
{
	scanf("%d %d",&n,&m);
	all=1<<n;
	int  u,v;
	for(int i=1;i<=m;i++)
	{
		scanf("%d %d",&u,&v);
		cnt[u][1<<(v-1)]=1;
	}
	for(int i=1;i<=n;i++)
		for(int s=1;s<all;s++)
		{
			int k=LowBit(s);
			cnt[i][s]=cnt[i][s-k]+cnt[i][k];
			p[i][s]=(1<<cnt[i][s])-1;
		}
	for(int s=0;s<all;s++)
		if((s&1)==((s&2)/2))
			legal[s]=true;
	dp[0]=1;
	for(int s=1;s<all;s++)
		if(legal[s])
		{
			dp[s]=1;
			for(int t=((s-1)&s);t;t=((t-1)&s))
				if(legal[t])
				{
					int res=s-t,add=dp[res];
					for(int i=0;i<n;i++)
						if((1<<i)&res)
							add=1LL*add*p[i+1][t]%MO;
						else if((1<<i)&t)
							add=1LL*add*(1LL<<cnt[i+1][res])%MO;
					dp[s]=(1LL*dp[s]+1LL*add)%MO;
				}
		}
	int ans=PowMod(2,m);
	ans=((1LL*ans-1LL*dp[all-1])+MO)%MO;
	printf("%d\n",ans);
	return 0;
}