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 <bits/stdc++.h>
#define sz(c)      int(c.size())
#define rep(i,a,b) for (int i=a; i<(b); ++i)
#define per(i,a,b) for (int i=(b)-1; i>=(a); --i)
#define bit(mask,i) ((mask)>>(i)&1)
using namespace std;
using ll = long long;


const int MD=int(1e9)+7;

inline int madd(int x,int y) {
	if (x+y<MD) return x+y; else return x+y-MD;
}
inline int msub(int x,int y) {
	if (x-y>=0) return x-y; else return x-y+MD;
}
inline int mmul(int x,int y) {
	return int(ll(x)*y % MD);
}

int const MAXN=16;
int N,M;
int a[MAXN];
int f[1<<MAXN];

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	cin>>N>>M;
	rep(i,0,M) {
		int x,y;
		cin>>x>>y;
		x--,y--;
		a[x]|=1<<y;
	}
	
	rep(mask,1,1<<N) if (bit(mask,0)==bit(mask,1)) {
		f[mask]=1;
		for (int S=mask&(mask-1); S>0; S=mask&(S-1)) {
			if (bit(S,0)!=bit(S,1)) continue;
			int T=mask^S;
			int add=f[S];
			assert(T!=0);
			rep(v,0,N) {
				if (bit(S,v)) {
					int k=__builtin_popcount(a[v]&T);
					add=mmul(add,(1<<k)-1);
				} else if (bit(T,v)) {
					int k=__builtin_popcount(a[v]&S);
					add=mmul(add,1<<k);
				}
			}
			f[mask]=madd(f[mask],add);
		}
	}
	
	int res=1;
	rep(i,0,M) res=madd(res,res);
	res=msub(res,f[(1<<N)-1]);
	cout<<res<<"\n";
}