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<cstdlib>
#include<cstring>
#define llong long long
#define ldouble long double
#define uint unsigned int
#define ullong unsigned long long
#define udouble unsigned double
#define uldouble unsigned long double
#define modinc(x) {if(x>=P) x-=P;}
#define pii pair<int,int>
#define piii pair<pair<int,int>,int>
#define piiii pair<pair<int,int>,pair<int,int> >
#define pli pair<llong,int>
#define pll pair<llong,llong>
#define Memset(a,x) {memset(a,x,sizeof(a));}
#define U ((1<<n)-1)
using namespace std;

const int N = 15;
const int P = 1e9+7;
int g[N+2][(1<<N)+2];
int cnt[N+2][(1<<N)+2];
llong dp[(1<<N)+2];
int n,m;

int main()
{
	scanf("%d%d",&n,&m);
	llong tot = 1ll;
	for(int i=1; i<=m; i++) tot = (tot<<1)%P;
	for(int i=1; i<=m; i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		x--; y--; cnt[x][1<<y] = 1;
	}
	for(int i=0; i<n; i++)
	{
		cnt[i][0] = 0;
		for(int j=1; j<=U; j++)
		{
			int lb = j&(-j);
			cnt[i][j] = cnt[i][j^lb]+cnt[i][lb];
//			printf("cnt%d %d %d\n",i,j,cnt[i][j]);
		}
	}
	dp[0] = 1ll; dp[3] = 1ll;
	for(int i=4; i<=U; i++)
	{
		if((i&3)==3 || (i&3)==0)
		{
			dp[i] = 1ll;
			for(int j=i; j>0; j=(j-1)&i)
			{
				if(j==i) continue;
				if((j&3)==3 || (j&3)==0)
				{
					llong cur = dp[j];
					for(int k=0; k<n; k++)
					{
						if(j&(1<<k))
						{
							cur *= (1ll<<cnt[k][i^j])-1ll; cur %= P;
						}
						else if(i&(1<<k))
						{
							cur *= (1ll<<cnt[k][j]); cur %= P;
						}
					}
					dp[i] += cur; modinc(dp[i]);
				}
			}
//			printf("%d %lld\n",i,dp[i]);
		}
	} printf("%lld\n",(tot-dp[U]+P)%P);
	return 0;
}