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 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define re register
#define inc(i,j,k) for(re int i=j;i<=k;i++)
#define ll long long
using namespace std;
const int N=305;
const int mod=1e9+7;
inline int read(){
	int x=0;
	char ch=getchar();
	while(ch<'0'||ch>'9') ch=getchar();
	while('0'<=ch&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x;
}
int n,m;
int lim[N][4];
ll f[N][N][N],ans;
int main(){
	n=read(),m=read();
	inc(i,1,m){
		int x=read(),y=read(),c=read()-1;
		if(!c){
			if(!lim[y][0]) lim[y][0]=x;
			else lim[y][0]=min(lim[y][0],x);
		}
		if(c==1){
			if(!lim[y][1]) lim[y][1]=x;
			else lim[y][1]=min(lim[y][1],x);
			if(!lim[y][2]) lim[y][2]=x;
			else lim[y][2]=max(lim[y][2],x);
		} 
		if(c==2){
			if(!lim[y][3]) lim[y][3]=x;
			else lim[y][3]=max(lim[y][3],x);
		}
	}
	f[0][0][0]=1;
	inc(i,0,n){
		inc(j,0,max(0,i-1)){
			inc(k,0,max(0,i-2)){
				 if(lim[i][0]&&(j>=lim[i][0]||k>=lim[i][0])) f[i][j][k]=0;
				 if(lim[i][1]&&j>=lim[i][1]&&k>=lim[i][1]) f[i][j][k]=0;
				 if(lim[i][2]&&j<lim[i][2]&&k<lim[i][2]) f[i][j][k]=0;
				 if(lim[i][3]&&(j<lim[i][3]||k<lim[i][3])) f[i][j][k]=0;
				 (f[i+1][j][k]+=f[i][j][k])%=mod;
				 (f[i+1][i][k]+=f[i][j][k])%=mod;
				 (f[i+1][i][j]+=f[i][j][k])%=mod;
			}
		}
	}
	inc(j,0,n-1){
		inc(k,0,n-2) (ans+=f[n][j][k])%=mod;
	}
	printf("%lld",ans);
}