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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <cmath>
#include <iomanip>
#include <numeric>
#include <climits>
#include <cstring>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll MOD = 1e9 + 7;

int main()
{
    int n, m, le[310], ri[310], x[310];

    ll dp[310][310][310];

    vector<P> v[310];

    cin >> n >> m;
    REP(i, m)
    {
        cin >> le[i] >> ri[i] >> x[i];
        v[ri[i]].push_back({le[i], x[i]});
    }

    memset(dp, 0, sizeof(dp));

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

    ll res = 0;

    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n; j++)
            for (int k = 0; k <= n; k++)
            {
                int ma = max(i, max(j, k));

                int a[3];
                a[0] = i;
                a[1] = j;
                a[2] = k;

                sort(a, a + 3);

                bool check = true;

                for (P l : v[ma])
                {
                    switch (l.second)
                    {
                    case 1:
                        if (l.first <= a[1])
                            check = false;
                        break;
                    case 2:
                        if (a[1] < l.first || a[0] >= l.first)
                            check = false;
                        break;
                    case 3:
                        if (a[0] < l.first)
                            check = false;
                        break;
                    default:
                        break;
                    }
                }

                if (!check)
                    dp[i][j][k] = 0;

                if (ma == n)
                {
                    res += dp[i][j][k];
                    res %= MOD;
                }

                (dp[ma + 1][j][k] += dp[i][j][k]) %= MOD;
                (dp[i][ma + 1][k] += dp[i][j][k]) %= MOD;
                (dp[i][j][ma + 1] += dp[i][j][k]) %= MOD;
            }

    cout << res << endl;

    return 0;
}