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 <bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for (int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

#define ctz __builtin_ctz
#define bcnt __builtin_popcount

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;

const int oo = 0x3f3f3f3f;

const int Mod = 1e9 + 7;

const int maxn = 15;

int dp[(1 << maxn) + 5];

int n, m;
int adj[maxn + 5];

int main()
{
#ifdef matthew99
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	scanf("%d%d", &n, &m);
	REP(i, 0, m)
	{
		int u, v;
		scanf("%d%d", &u, &v), --u, --v;
		adj[u] |= 1 << v;
	}
	dp[0] = 1;
	REP(i, 0, 1 << n) if (dp[i])
	{
		int res = ((1 << n) - 1) ^ i;
		for (int j = res; j > 0; (--j) &= res)
		{
			if ((j & 3) == 3) continue;
			int val = dp[i];
			REP(k, 0, n) if (!(i >> k & 1))
			{
				if (j >> k & 1) val = ((LL)val << bcnt(adj[k] & ~(i | j))) % Mod;
				else val = (LL)val * ((1 << bcnt(adj[k] & j)) - 1) % Mod;
			}
			(dp[i | j] += val) %= Mod;
		}
	}
	int ans = dp[(1 << n) - 1];
	(ans += Mod) %= Mod;
	printf("%d\n", ans);
	return 0;
}
