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>
typedef long long int64;
static const int MAXN = 15;
static const int MODULUS = 1e9 + 7;
#define _  %  MODULUS
#define __ %= MODULUS

int n, m;
int out[MAXN] = { 0 };
int64 f[1 << MAXN];

int main()
{
    scanf("%d%d", &n, &m);
    int u, v;
    for (int i = 0; i < m; ++i) {
        scanf("%d%d", &u, &v); --u, --v;
        out[u] |= (1 << v);
    }

    f[0] = 1;
    for (int s = 1; s < (1 << n); ++s) if (!!(s & 1) == !!(s & 2)) {
        for (int t = s; t > 0; t = (t - 1) & s) if (!!(t & 1) == !!(t & 2)) {
            int64 cur = f[s ^ t];
            for (int u = 0; u < n; ++u) if (s & (1 << u)) {
                (cur *= (t & (1 << u)) ?
                    (1LL << __builtin_popcount(out[u] & (s ^ t)))_ :
                    ((1LL << __builtin_popcount(out[u] & t)) - 1)_)__;
            }
            (f[s] += cur)__;
        }
    }

    int64 p = 1;
    for (int i = 0; i < m; ++i) (p += p)__;
    printf("%lld\n", (p - f[(1 << n) - 1] + MODULUS)_);
    return 0;
}