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 N = 310, MOD = 1e9 + 7;
struct data {
  int l,x;
};
vector<data> req[N];
int dp[N][N][N],n,m,ans;
int main() {
  int x,y,z;
  scanf("%d%d",&n,&m);
  for (int i = 1 ; i <= m ; ++ i) {
    scanf("%d%d%d",&x,&y,&z);
    req[y].push_back((data){x,z});
  }
  dp[0][0][1] = 1;
  dp[0][1][0] = 1;
  dp[1][0][0] = 1;
  for (int k = 1 ; k <= n ; ++ k) {
    for (int i = 0 ; i < (int)req[k].size() ; ++ i)
      for (int a = 0 ; a < k ; ++ a)
        for (int b = 0 ; b < k ; ++ b) if (a != b || (!a)) {
          int cur = 1;
          cur += (a >= req[k][i].l);
          cur += (b >= req[k][i].l);
          if (cur != req[k][i].x) {
            dp[k][a][b] = dp[a][k][b] = dp[a][b][k] = 0;
          }
        }
    if (k == n) break;
    for (int i = 0 ; i < k ; ++ i)
      for (int j = 0 ; j < k ; ++ j)
        if (i != j || (!i)) {
          #define _(x) ((x) %= MOD)
          _(dp[k+1][i][j] += dp[k][i][j]);
          _(dp[k][k+1][j] += dp[k][i][j]);
          _(dp[k][i][k+1] += dp[k][i][j]);
          _(dp[k+1][k][j] += dp[i][k][j]);
          _(dp[i][k+1][j] += dp[i][k][j]);
          _(dp[i][k][k+1] += dp[i][k][j]);
          _(dp[k+1][j][k] += dp[i][j][k]);
          _(dp[i][k+1][k] += dp[i][j][k]);
          _(dp[i][j][k+1] += dp[i][j][k]);
        }
  }
  for (int i = 0 ; i < n ; ++ i)  
    for (int j = 0 ; j < n ; ++ j)
      if (i != j || (!i)) {
        _(ans += dp[n][i][j]);
        _(ans += dp[i][n][j]);
        _(ans += dp[i][j][n]);
      }
  printf("%d\n",ans);
  return 0;
}