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>
typedef long long i64;
const int P = 1e9 + 7;
int N, M;
int dp[1 << 15];
int ouset[1 << 15];

int main() {
  if(fopen("input.txt","r")) {
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
  }
  
  scanf("%d%d", &N, &M);
  while(M--) {
    int u, v; scanf("%d%d", &u, &v);
    --u, --v; ouset[u] |= 1 << v;
  }

  dp[0] = 1;
  for(int S = 0;S < 1<<N; ++S) {
    if(!dp[S]) continue;
    int T = ((1 << N) - 1) ^ S;
    for(int S0 = T; S0; (--S0) &= T) {
      if((S0 & 3) == 3) continue;
      int w = dp[S];
      for(int k = 0;k < N; ++k) {
	if(S >> k & 1) continue;
	if(S0 >> k & 1) w = (i64) w * (1ll << __builtin_popcount(ouset[k] & ~(S | S0))) % P;
	else w = (i64) w * ((1 << __builtin_popcount(ouset[k] & S0)) - 1) % P;
      }
      (dp[S | S0] += w) %= P;
    }
  }

  printf("%d\n", dp[(1 << N) - 1]);
  
  return 0;
}