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>

using namespace std;

const int Max_N(15);
typedef long long int LL;
const int MOD(1000000000 + 7);

inline void update(int &a, const int &b)
{
	((a += b) >= MOD) ? (a -= MOD) : 0;
}

inline int Mult(const int &a, const int &b)
{
	return (a * 1LL) * b % MOD;
}

int N, M, F[1 << Max_N], power[Max_N * Max_N], In[Max_N][1 << Max_N];
bool G[Max_N][Max_N];

//注意到，如果我们把所有点按照其SG值分层，那么一定会分成0 ~ max{SG(u)}这么多层
//每个点一定会向前面的每一层中的点都至少连一条边，同时每个点不会向同一层中的点连边 
//考虑状压dp计数，每次枚举SG值最小的一层。转移时1和2不在同一层 

void init()
{
	scanf("%d%d", &N, &M);
	for (int i = 1, s, t;i <= M;++i)
		scanf("%d%d", &s, &t), --s, --t, G[s][t] = true;
	power[0] = 1;
	for (int i = 1;i <= M;++i)
		power[i] = Mult(power[i - 1], 2);
	for (int t = 0;t <= N - 1;++t)
		for (int S = 0;S < (1 << N);++S)
			for (int s = 0;s <= N - 1;++s)
				if (S & (1 << s))
					In[t][S] += G[t][s];
}

void dp()
{
	F[0] = 1;
	for (int S = 1;S < (1 << N);++S)
		for (int S0 = S, val;S0;S0 = ((S0 - 1) & S))
		{
			if (((S0 & (1 << 0)) != 0) && ((S0 & (1 << 1)) != 0))
				continue;
			val = F[S - S0];
			for (int i = 0;i <= N - 1;++i)
				if ((S - S0) & (1 << i))
					val = Mult(val, ((power[In[i][S0]] - 1) % MOD + MOD) % MOD);
			for (int i = 0;i <= N - 1;++i)
				if (S0 & (1 << i))
					val = Mult(val, power[In[i][S - S0]]);
			update(F[S], val);
		}
	printf("%d", F[(1 << N) - 1]);
}

int main()
{
	init();
	dp();
	return 0;
}