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 fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long ll;

const int N=20,S=(1<<15)+5,Mo=1e9+7;

int n,m,x,y,cnt[N][S],p[N*N],f[S];
bool ok[S];

int main() {
	scanf("%d%d",&n,&m);
	p[0]=1;
	fo(i,1,m) {
		scanf("%d%d",&x,&y);
		fo(s,0,(1<<n)-1) if (s&(1<<y-1)) cnt[x][s]++;
		p[i]=(p[i-1]<<1)%Mo;
	}
	f[0]=1;
	fo(s,0,(1<<n)-1) if (((s&1)>0)==((s&2)>0)) ok[s]=1;
	fo(s,1,(1<<n)-1)
		if (ok[s]) {
			f[s]=1;
			for(int t=(s-1)&s;t;t=(t-1)&s) 
				if (ok[t]) {
					int ss=s^t,now=f[ss];
					fo(i,1,n) {
						if (t&(1<<(i-1))) now=(ll)now*p[cnt[i][ss]]%Mo;
						if (ss&(1<<(i-1))) now=(ll)now*(p[cnt[i][t]]-1)%Mo;
					}
					(f[s]+=now)%=Mo;
				}
		}
	printf("%d\n",(p[m]-f[(1<<n)-1]+Mo)%Mo);
	return 0;
}