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
#include <bits/stdc++.h>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

typedef long long ll;
const int MAXM = (1 << 15) + 5, MAXN = 20;
const int mod = 1e9 + 7;
 
inline void inc(int &x, int y) { x += y; if (x >= mod) x -= mod; }
 
int f[MAXM], bit2[MAXM], n, G[MAXN], m;
 
int main()
{
	cin >> n >> m;
	bit2[0] = 1; f[0] = 1;
	for (int i = 1; i <= m; i ++) bit2[i] = (bit2[i - 1] << 1) % mod;
	for (int i = 1, x, y; i <= m; i ++) cin >> x >> y, --x, --y, G[x] |= 1 << y;
	for (int S = 1; S < (1 << n); S ++) if ((S & 1) == (S >> 1 & 1))
		for (int U = S; U; U = U - 1 & S) if ((U & 1) == (U >> 1 & 1))
		{
			int ways = 1;
			ll T = (S ^ U);
			for (int i = 0; i < n; i ++) if (S >> i & 1)
			{
				if (U >> i & 1) ways = 1LL * ways * bit2[__builtin_popcount(G[i] & T)] % mod;
				else ways = 1LL * ways * (bit2[__builtin_popcount(G[i] & U)] - 1) % mod;
			}
			f[S] = (f[S] + 1LL * f[T] * ways % mod) % mod;
		}
	return printf("%d\n", (bit2[m] - f[(1 << n) - 1] + mod) % mod), 0;
}