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 xx first
#define yy second
#define mp make_pair
#define pb push_back
#define fill(x, y) memset(x, y, sizeof x)
#define copy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair < int, int > pa;

inline int read()
{
	int sc = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') sc = sc * 10 + ch - '0', ch = getchar();
	return sc * f;
}

const int MAXN = 15;
const int MAXM = (1 << 15) + 5;
const int mod = 1e9 + 7;

inline void inc(int &x, int y) { x += y; if (x >= mod) x -= mod; }

int f[MAXM], table[MAXM], n, G[MAXN], m;

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = read(), m = read();
	table[0] = 1; f[0] = 1;
	for (int i = 1; i <= m; i ++) table[i] = (table[i - 1] << 1) % mod;
	for (int i = 1, x, y; i <= m; i ++) x = read() - 1, y = read() - 1, G[x] |= 1 << y;
	for (int S = 1; S < (1 << n); S ++) if ((S & 1) == (S >> 1 & 1))
		for (int T = S; T; T = T - 1 & S) if ((T & 1) == (T >> 1 & 1))
		{
			int ways = 1;
			for (int i = 0; i < n; i ++) if (S >> i & 1)
			{
				if (T >> i & 1) ways = 1LL * ways * table[__builtin_popcount(G[i] & (S ^ T))] % mod;
				else ways = 1LL * ways * (table[__builtin_popcount(G[i] & T)] - 1) % mod;
			}
			inc(f[S], 1LL * f[S ^ T] * ways % mod);
		}
	return printf("%d\n", (table[m] - f[(1 << n) - 1] + mod) % mod), 0;
}