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 <bits/stdc++.h>
#define EB emplace_back

typedef std::pair <int, int> pr;
typedef std::vector <pr> vecpr;
const int N = 324, mod = 1000000007;

int n, q;
int dp[2][N][N], (*cur)[N] = *dp, (*nxt)[N] = dp[1];
vecpr lim[N];

inline void add(int &x, const int y) {x += y - mod, x += x >> 31 & mod;}

int main() {
	int i, j, k, l, r, x, ans = 0;
	scanf("%d%d", &n, &q);
	for (; q; --q) scanf("%d%d%d", &l, &r, &x), lim[r].EB(l, x);
	**nxt = 1;
	for (i = 1; i <= n; ++i) {
		std::swap(cur, nxt), memset(nxt, 0, sizeof *dp);
		for (const pr &w : lim[i]) 
			switch (std::tie(l, x) = w, x) {
				case 1: for (j = l; j < i; ++j) memset(cur[j], 0, (j + 1) << 2); break;
				case 2:
					for (j = 0; j < l; ++j) memset(cur[j], 0, (j + 1) << 2);
					for (; j < i; ++j) memset(cur[j] + l, 0, (j - l + 1) << 2); break;
				case 3: for (j = 0; j < i; ++j) memset(cur[j], 0, l << 2); break;
			}
		if (i == n) break;
		for (j = 0; j < i; ++j)
			for (k = 0; !k || k < j; ++k) if ((x = cur[j][k]))
				add(nxt[j][k], x), add(nxt[i][k], x), add(nxt[i][j], x);
	}
	for (j = 0; j < n; ++j) ans = (ans + std::accumulate(cur[j], cur[j] + (j + 1), 0ll)) % mod;
	printf("%d\n", 3ll * ans % mod);
	return 0;
}