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;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second

template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, pii> P;

const ll INF = 1ll<<30;
const ll MOD = 1000000007;
const double EPS  = 1e-10;

int N, M;
vector<pii> lv[301]; // [r_i] = (l_i, x_i)
ll dp[301][301][301];

int main() {
	cin >> N >> M;
	REP(i, M) {
		int l, r, x;
		scanf("%d %d %d", &l, &r, &x);
		lv[r].push_back(pii(l, x));
	}
	
	ll ans = 0;
	
	dp[0][0][0] = 1;
	REP(i, N + 1) {
		REP(j, N + 1) {
			REP(k, N + 1) {
				int r = max(i, max(j, k));
			
				REP(l, lv[r].size()) {
					int cnt = (i >= lv[r][l].fi) + (j >= lv[r][l].fi) + (k >= lv[r][l].fi);
					if (cnt != lv[r][l].se) dp[i][j][k] = 0;
				}
				
				if (dp[i][j][k] == 0) continue;
				
				if (r == N) {
					(ans += dp[i][j][k]) %= MOD;
					continue;
				}
				
				(dp[r + 1][j][k] += dp[i][j][k]) %= MOD;
				(dp[i][r + 1][k] += dp[i][j][k]) %= MOD;
				(dp[i][j][r + 1] += dp[i][j][k]) %= MOD;
			}
		}
	}
	
	cout << ans << endl;
	
	return 0;
}