Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

3 1
1 3 3

Sample Output 1

6

The six ways are:

RGB

RBG

GRB

GBR

BRG

BGR

where R, G and B correspond to red, green and blue squares, respectively.

Sample Input 2

4 2
1 3 1
2 4 2

Sample Output 2

6

The six ways are:

RRRG

RRRB

GGGR

GGGB

BBBR

BBBG

Sample Input 3

1 3
1 1 1
1 1 2
1 1 3

Sample Output 3

0

There are zero ways.

Sample Input 4

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

Sample Output 4

108

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define mp make_pair
#define pb push_back

#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, n) for (int i = 0; i < (int)(n); ++ i)

const int mod = 1e9 + 7;
int n, q;
int lim[305][305];
int dp[305][305][305];

int f(int i, int a, int b, int c) {
	for (int l = 1; l <= i; ++ l) if (~lim[l][i]) {
		int cnt = (a >= l) + (b >= l) + (c >= l);
		if (cnt != lim[l][i]) return dp[a][b][c] = 0;
	}
	if (~dp[a][b][c]) return dp[a][b][c];
	if (i == n) return 1;
	return dp[a][b][c] = (0LL + f(i + 1, i + 1, b, c) + f(i + 1, a, i + 1, c) + f(i + 1, a, b, i + 1)) % mod;
}

int main() {
	scanf("%d %d", &n, &q);
	memset(lim, -1, sizeof(lim));
	rep(i, q) {
		int l, r, x;
		scanf("%d %d %d", &l, &r, &x);
		if (~lim[l][r] && lim[l][r] == x) { puts("0"); return 0; }
		lim[l][r] = x;
	}
	memset(dp, -1, sizeof(dp));
	printf("%d\n", f(0, 0, 0, 0));
	return 0;
}