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: Wrong Answer)

//https://www.cnblogs.com/yoyoball/p/9476080.html
//copy from yoyoball

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#define Pr pair<int,int>
#define mp make_pair
using namespace std;
const int N=310,MOD=1e9+7;
vector<Pr> rec[N];
int f[N][N][N];
int n,m,ans;
void update(int &x,int y){x=(1LL*x+y+MOD+MOD)%MOD;}
void solve(){
    int sz,l,x;
    f[1][0][0]=1;
    for (int i=1;i<=n;++i){
        sz=rec[i].size();
        for (int p=0;p<sz;++p){
            l=rec[i][p].first; x=rec[i][p].second;
            for (int j=0;j<i;++j)
                for (int k=0;k<=max(0,j-1);++k){
                    if (x==1){
                        if (l<=j) f[i][j][k]=0;
                    }
                    else if (x==2){
                        if (l<=k||j<l) f[i][j][k]=0;
                    }
                    else{
                        if (k<l) f[i][j][k]=0;
                    }
                }
        }
        if (i==n) break;
        for (int j=0;j<i;++j)
            for (int k=0;k<=max(j-1,0);++k){
                if (!f[i][j][k]) continue;
                update(f[i+1][j][k],f[i][j][k]);
                update(f[i+1][i][k],f[i][j][k]);
                update(f[i+1][i][j],f[i][j][k]);
            }
    }
    ans=0;
    for (int j=0;j<n;++j)
        for (int k=0;k<=max(0,j-1);++k)
            update(ans,f[n][j][k]);
    ans=1LL*ans*3%MOD;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
#endif
    int l,r,x;
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;++i){
        scanf("%d%d%d",&l,&r,&x);
        rec[r].push_back(mp(l,x));
    }
    solve();
    printf("%d\n",ans);
}