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 <cstdio>
#include <cstring>
#include <algorithm>
#define P 1000000007
#define ct(x) __builtin_popcount(x)
#define Log(x) (Sym[(x) % 37])

using namespace std;

int n, m, B[15], R[15], Sym[55];

long long F[1 << 15], ans = 1;

int main() {
    for(int i = 0; i < 30; i++) Sym[(1 << i) % 37] = i;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) ans = ans * 2 % P;
    while(m--) {
        int x, y;
        scanf("%d%d", &x, &y);
        B[--x] |= (1 << --y);
        R[y] |= (1 << x);
    }
    F[0] = 1;
    for(int i = 1; i < (1 << n); i++) if(!!(i & 1) ^ !(i & 2))
        for(int j = i; j; j = (j - 1) & i) if(!!(j & 1) ^ !(j & 2)) {
            long long s = (1 << n) - i - 1, tmp = 1;
            for(int k = 0; k < n; k++)
                if((s >> k) & 1) tmp = tmp * ((1 << ct(B[k] & j)) - 1) % P * (1 << ct(R[k] & j)) % P;
            F[i] = (F[i - j] * tmp + F[i]) % P;
        }
    printf("%lld\n", ((ans - F[(1 << n) - 1]) % P + P) % P);
    return 0;
}