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<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
#include<cctype>
#define pa pair<int,int>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define mp make_pair
#define ll long long
#define ull unsigned long long
#define pb push_back

using namespace std;

inline ll read()
{
	long long f=1,sum=0;
	char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-1;c=getchar();}
	while (isdigit(c)) {sum=sum*10+c-'0';c=getchar();}
	return sum*f;
}
const int MAXN=20;
const int MAXM=(1<<15)+10; 
#define Mod 1000000007
int f[MAXM],P[MAXN];
int e[MAXN];
ll quickpow(ll a,int b)
{
	ll ret=1;
	while (b)
	{
		if (b&1) ret*=a,ret%=Mod;
		b/=2,a*=a,a%=Mod;
	}
	return ret;
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	P[0]=1;
	for (int i=1;i<=n;i++)
		P[i]=(P[i-1]*2)%Mod;
	for (int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);u--,v--;
		e[u]|=(1<<v);
	}
	f[0]=1;
	for (int s=1;s<1<<n;s++)
	{
		//if ((s&1)!=((s&2)>>1)) continue;
		if ((s&1)!=(s>>1&1)) continue;
		for (int t=s;t;t=(t-1)&s)
		{
			if ((t&1)!=(t>>1&1)) continue;
			//if ((t&1)!=((t&2)>>1)) continue;
			int tmp=1;
			for (int i=0;i<n;i++)
			{
				if (!(s&(1<<i))) continue;
				if (t&(1<<i)) tmp=1ll*tmp*P[__builtin_popcount(e[i]&(s^t))]%Mod;
				else tmp=1ll*tmp*(P[__builtin_popcount(e[i]&t)]-1)%Mod;
			}
			f[s]+=1ll*tmp*f[s^t]%Mod;
			f[s]%=Mod;
		}
	}
	ll ans=0;
	ans=quickpow(2,m)-f[(1<<n)-1]+Mod;
	ans%=Mod;
	cout<<ans;
	return 0;
}