Problem: There are N squares arranged in a row. The squares are numbered 1, 2, ..., N, from left to right.

Snuke is painting each square in red, green or blue. According to his aesthetic sense, the following M conditions must all be satisfied. The i-th condition is:

* There are exactly x_i different colors among squares l_i, l_i + 1, ..., r_i.



In how many ways can the squares be painted to satisfy all the conditions? Find the count modulo 10^9+7.

Constraints

* 1 ≤ N ≤ 300
* 1 ≤ M ≤ 300
* 1 ≤ l_i ≤ r_i ≤ N
* 1 ≤ x_i ≤ 3

Input

Input is given from Standard Input in the following format:


N M
l_1 r_1 x_1
l_2 r_2 x_2
:
l_M r_M x_M


Output

Print the number of ways to paint the squares to satisfy all the conditions, modulo 10^9+7.

Examples

Input

3 1
1 3 3


Output

6


Input

4 2
1 3 1
2 4 2


Output

6


Input

1 3
1 1 1
1 1 2
1 1 3


Output

0


Input

8 10
2 6 2
5 5 1
3 5 2
4 7 3
4 4 1
2 3 1
7 7 1
1 5 2
1 7 3
3 4 2


Output

108

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 300, MOD = 1e9 + 7;

inline void Add(int &x, int y) {
	(x += y) >= MOD && (x -= MOD);
}

int n, m, f[N + 2][N + 2][N + 2];
vector<pair<int, int>> lim[N + 5];

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1, l, r, x; i <= m; ++i) {
		scanf("%d%d%d", &l, &r, &x);
		lim[r].push_back(make_pair(l, x));
	}
	f[1][0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (const auto &p : lim[i]) {
			int l = p.first, x = p.second;
			for (int j = 0; j < i; ++j) {
				for (int k = 0; k <= max(j - 1, 0); ++k) {
					if (x == 1 && j >= l) f[i][j][k] = 0;
					if (x == 2 && (k >= l || j < l)) f[i][j][k] = 0;
					if (x == 3 && k < l) f[i][j][k] = 0;
				}
			}
		}
		for (int j = 0; j < i; ++j) {
			for (int k = 0; k <= max(j - 1, 0); ++k) {
				Add(f[i + 1][j][k], f[i][j][k]);
				Add(f[i + 1][i][k], f[i][j][k]);
				Add(f[i + 1][i][j], f[i][j][k]);
			}
		}
	}
	int ans = 0;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j <= max(i - 1, 0); ++j) {
			Add(ans, f[n][i][j]);
		}
	}
	printf("%lld\n", 3LL * ans % MOD);
	return 0;
}