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<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;

int n,m,ans;
int f[310][310][310];

struct node{
	int l,x;
};
vector<node>e[310];

void read(int &x)
{
    x=0;int f=1;char s=getchar();
    while(s<'0'||s>'9') {if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9') {x=x*10+s-'0';s=getchar();}
    x*=f;
}

bool check(int i,int j,int k)
{
	int r=max(i,max(j,k)),l,x,cnt=0;
	for(int u=0;u<e[r].size();u++)
	{
		l=e[r][u].l;
		x=e[r][u].x;
		cnt=(i>=l)+(j>=l)+(k>=l);
		if(cnt!=x) return 0;
	}
	return 1;
}

int main()
{
    read(n);read(m);
    int l,r,x;
    for(int i=1;i<=m;i++)
    {
    	read(l);read(r);read(x);
    	e[r].push_back((node){l,x});
    }
    f[0][0][0]=1;
    for(int i=0;i<=n;++i)
    	for(int j=0;j<=n;++j)
    		for(int k=0;k<=n;++k)
    		if(f[i][j][k])
    		{
    			if((i&&j&&i==j)||(i&&k&&i==k)||(j&&k&&j==k)) continue;
    			if(!check(i,j,k)) continue;
    			r=max(i,max(j,k));
    			f[r+1][j][k]=(f[r+1][j][k]+f[i][j][k])%MOD;
    			f[i][r+1][k]=(f[i][r+1][k]+f[i][j][k])%MOD;
    			f[i][j][r+1]=(f[i][j][r+1]+f[i][j][k])%MOD;
    			if(r==n) ans=(ans+f[i][j][k])%MOD;
    		}
    printf("%d\n",ans);
    return 0;
}