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 PYTHON3):

Solution:
import sys
from collections import defaultdict

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]

# 一番後ろの色を(赤に)固定した上で、もらうDPに書き換え
def main():
    md = 10 ** 9 + 7
    n, m = MI()
    rtol = defaultdict(list)
    for _ in range(m):
        l, r, x = MI()
        rtol[r].append([l, x])
    dp = [[0] * (n + 1) for _ in range(n + 1)]
    dp[0][0] = 1
    for k in range(1, n + 1):
        for i in range(k - 1):
            dp[i][k - 1] = dp[k - 1][i] = sum(dp[i][:k - 1]) % md
        for l, x in rtol[k]:
            if x == 1:
                for i in range(l, k):
                    for j in range(k):
                        dp[i][j] = dp[j][i] = 0
            if x == 2:
                for i in range(l, k):
                    for j in range(i + 1, k):
                        dp[i][j] = dp[j][i] = 0
                for i in range(l):
                    for j in range(i + 1):
                        dp[i][j] = dp[j][i] = 0
            if x == 3:
                for i in range(l):
                    for j in range(k):
                        dp[i][j] = dp[j][i] = 0

    # p2D(dp)
    print(sum(sum(dr) for dr in dp) * 3 % md)

main()