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 <bits/stdc++.h>
using namespace std;

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

const ll MAX=305;
const ll INF=0x3f3f3f3f;
const ll MOD=1e9+7;
struct range{
    ll x,y,c;
};
ll dp[2][MAX][MAX],ans,n,m;
range cst[MAX];
vector<ll> add[MAX],sub[MAX];
int main(){
    scanf("%lld%lld",&n,&m);
    for(ll i=0;i<m;i++){
        scanf("%lld%lld%lld",&cst[i].x,&cst[i].y,&cst[i].c);
        add[cst[i].x].push_back(i);
        sub[cst[i].y].push_back(i);
    }
    dp[0][0][0]=1;
    for(ll i=1;i<=n;i++){
        for(ll j=0;j<i;j++) for(ll k=0;k<=j;k++) dp[i%2][j][k]=0;
        for(ll j=0;j<i;j++) for(ll k=0;k<=j;k++){
            dp[i%2][i-1][k]=(dp[i%2][i-1][k]+dp[(i+1)%2][j][k])%MOD;
            dp[i%2][i-1][j]=(dp[i%2][i-1][j]+dp[(i+1)%2][j][k])%MOD;
            dp[i%2][j][k]=(dp[i%2][j][k]+dp[(i+1)%2][j][k])%MOD;
        }
        for(ll j=0;j<i;j++) for(ll k=0;k<=j;k++){
            for(auto vec:sub[i]){
                if(cst[vec].c!=1+(cst[vec].x<=j)+(cst[vec].x<=k)) dp[i%2][j][k]=0;
            }
        }
    }
    for(ll j=0;j<n;j++) for(ll k=0;k<=j;k++) ans=(ans+dp[n%2][j][k])%MOD;
    printf("%lld\n",ans);
}