Problem: 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.

Examples

Input

3 1
1 3 3


Output

6


Input

4 2
1 3 1
2 4 2


Output

6


Input

1 3
1 1 1
1 1 2
1 1 3


Output

0


Input

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


Output

108

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <cstdio>
#include <ctime>
#include <assert.h>
#include <chrono>
#include <random>
#include <numeric>
#include <set>
using namespace std;
typedef long long int ll;
typedef unsigned long long ull;

constexpr ll mod=1e9+7;
vector<pair<int,int>> v[301];
ll dp[303][303]={1};

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n,m; cin >> n >> m;
    while(m--){
        int l,r,x; cin >> l >> r >> x;
        v[r-1].push_back({l,x});
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<i;j++){
            for(int k=0;k<i;k++){
                (dp[i][j]+=dp[max(j,k)][min(j,k)])%=mod;
            }
        }
        for(auto p:v[i]){
            for(int j=0;j<=i;j++){
                for(int k=0;k<=j;k++){
                    if(p.second+(j<p.first)+(k<p.first)-3){
                        dp[j][k]=0;
                    }
                }
            }
        }
    }
    ll res=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<i;j++){
            (res+=6*dp[i][j])%=mod;
        }
    }
    printf("%lld\n",(res+dp[0][0]*3)%mod);
}