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<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const long long MAXN=16,MOD=1e9+7;
long long n,m;
long long ts=0;
long long f[1<<MAXN],cnt[MAXN][1<<MAXN];
vector<int>w[MAXN];
int main()
{
	scanf("%lld %lld",&n,&m);
	for(long long i=1;i<=m;i++)
	{
		long long u,v;
		scanf("%lld %lld",&u,&v);
		w[u].push_back(v);
	}
	ts=(1<<n)-1;
	for(long long i=1;i<=n;i++)
	{
		for(long long S=0;S<=ts;S++)
		{
			for(long long j=0;j<int(w[i].size());j++)
			{
				long long v=w[i][j];
				if((1<<(v-1))&S)
				{
					cnt[i][S]++;
				}
			}
		}
	}
	for(long long S=0;S<=ts;S++)
	{
		if((S&1)==((S>>1)&1))
		{
			f[S]=1;
			for(long long jS=(S-1)&S;jS;jS=(jS-1)&S)
			{
				if((jS&1)==((jS>>1)&1))
				{
					long long bS=S-jS;
					long long inc=f[bS];
					for(long long i=1;i<=n;i++)
					{
						if(bS&(1<<(i-1)))
						{
							inc=inc*((1<<cnt[i][jS])-1)%MOD;
						}
						else if(jS&(1<<(i-1)))
						{
							inc<<=cnt[i][bS];
							inc%=MOD;
						}
					}
					f[S]=(f[S]+inc)%MOD;
				}
			}
		}
	}
	long long ans=1;
	for(long long i=1;i<=m;i++)
		ans=ans*2%MOD;
	ans-=f[ts];
	ans=(ans%MOD+MOD)%MOD;
	printf("%lld\n",ans);
}