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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef long long ll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define sz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef long double ldouble;
typedef pair<double, double> pdd;

const int MOD = 1e9 + 7;
ll D[1<<15];
int n, m, E[20][20];
int P[500];

inline int f(int a){return a == n-1 ? a-1 : a; }

void solve() {
	cin >> n >> m;
	P[0] = 1;
	for(int i=1;i<500;i++) P[i] = (P[i-1] + P[i-1]) % MOD;
	rep(i, m){
		int x, y; cin >> x >> y;
		x = n - x, y = n - y;
		E[x][y] = 1;
	}
	D[0] = 1;
	for(int i=1;i<1<<(n-1);i++) {
		for(int j=i;j;j=(j-1)&i){
			ll C = 1;
			for(int b=0;b<n;b++) if(!(1<<f(b) & j)) {
				int cnt = 0;
				for(int a=0;a<n;a++) if(1<<f(a) & j) {
					cnt += E[b][a];
				}
				if(!(1<<f(b) & i)) C = C * (P[cnt] + MOD - 1) % MOD;
				else C = C * P[cnt] % MOD;
			}
			D[i] = (D[i] + D[i-j] * C) % MOD;
		}
	}
	ll ans = P[m] - D[(1<<(n-1))-1];
	if(ans < 0) ans += MOD;
	printf("%lld\n", ans);
}

int main(){
	int Tc = 1; //scanf("%d\n", &Tc);
	for(int tc=1;tc<=Tc;tc++){
		solve();
	}
	return 0;
};