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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 300 + 5;

int dp[maxn][maxn][maxn];
vector<pair<int, int> > Op[maxn];

bool chk(int r, int g, int b) {
    int p = max(r, max(g, b));
    for(auto x : Op[p]) {
        int cnt = 0;
        if(x.first <= r) ++cnt;
        if(x.first <= g) ++cnt;
        if(x.first <= b) ++cnt;

        if(cnt != x.second) return false;
    }
    return true;
}
int main() {

    int n, m;
    //freopen("data.txt", "r", stdin);

    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; i++) {
        int l, r, x;
        cin >> l >> r >> x;
        Op[r].push_back(make_pair(l, x));
    }

    dp[0][0][0] = 1;

    int v, ans = 0;
    for(int i = 0; i <= n; i++) 
        for(int j = 0; j <= n; j++)
            for(int k = 0; k <= n; k++) if((v = dp[i][j][k]) && chk(i, j, k)) {

                int p = max(i, max(j, k));
                if(p == n) (ans += v) %= mod;
                else {
                    ++p;
                    (dp[p][j][k] += v) %= mod;
                    (dp[i][p][k] += v) %= mod;
                    (dp[i][j][p] += v) %= mod;
                }
            }
    printf("%d\n", ans);

    return 0;
}