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 rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
#define popc __builtin_popcount
using std::cin; using std::cout;
const int maxn = 15;
typedef long long ll;
const int mod = 1e9 + 7;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
inline void reduce(int & x) {
	x += x >> 31 & mod;
}
int n, m;
int edge[maxn];
int f[1 << maxn];
int shift[maxn * maxn];
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	shift[0] = 1;
	rep(i, 1, maxn * maxn - 1) {
		reduce(shift[i] = shift[i - 1] + shift[i - 1] - mod);
	}
	rep(i, 1, m) {
		int x, y;
		cin >> x >> y;
		-- x, -- y;
		edge[x] |= 1 << y;
	}
	* f = 1;
	for(int i = 2;i < 1 << n;i += 2) {
		for(int j = i;j;j = j - 1 & i) {
			int s0 = i | (i >> 1 & 1);
			int s1 = j | (j >> 1 & 1);
			s0 ^= s1;
			int mul = f[i ^ j];
			rep(i, 0, n - 1){
				if(s1 >> i & 1) mul = (ll) mul * shift[popc(edge[i] & s0)] % mod;
				if(s0 >> i & 1) mul = (ll) mul * (shift[popc(edge[i] & s1)] - 1) % mod;
			}
			reduce(f[i] += mul - mod);
		}
	}
	int ans = shift[m] - f[(1 << n) - 2];
	cout << ans + (ans >> 31 & mod) << '\n';
}