Problem: You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.

In each spot, if a house has height a, you will gain a^2 dollars from it.

The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.

You would like to build houses to maximize your profit. Determine the maximum profit possible.

Input

The first line contains three integers n, h, and m (1 ≤ n,h,m ≤ 50) — the number of spots, the maximum height, and the number of restrictions.

Each of the next m lines contains three integers l_i, r_i, and x_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ x_i ≤ h) — left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.

Output

Print a single integer, the maximum profit you can make.

Examples

Input


3 3 3
1 1 1
2 2 3
3 3 2


Output


14


Input


4 10 2
2 3 8
3 4 7


Output


262

Note

In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.

In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.

In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.

In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).

Solution (in PYTHON3):

Solution:
T_ON = 0
DEBUG_ON = 1
MOD = 998244353


def solve():
    n, h, m = read_ints()
    A = [h for _ in range(n)]
    for i in range(m):
        l, r, x = read_ints()
        for j in range(l - 1, r):
            A[j] = min(A[j], x)
    total = 0
    for a in A:
        total += a * a
    print(total)


def main():
    T = read_int() if T_ON else 1
    for i in range(T):
        solve()


def debug(*xargs):
    if DEBUG_ON:
        print(*xargs)


from collections import *
import math


#---------------------------------FAST_IO---------------------------------------

import os
import sys
from io import BytesIO, IOBase
# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#----------------------------------IO_WRAP--------------------------------------

def read_int():
    return int(input())


def read_ints():
    return list(map(int, input().split()))


def print_nums(nums):
    print(" ".join(map(str, nums)))


def YES():
    print("YES")


def Yes():
    print("Yes")


def NO():
    print("NO")


def No():
    print("No")


def First():
    print("First")


def Second():
    print("Second")


#----------------------------------FIB--------------------------------------

def fib(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
    return a


def fib_ns(n):
    assert n >= 1
    f = [0 for _ in range(n + 1)]
    f[0] = 0
    f[1] = 1
    for i in range(2, n + 1):
        f[i] = f[i - 1] + f[i - 2]
    return f

#----------------------------------MOD--------------------------------------

def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def xgcd(a, b):
    """return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
    x0, x1, y0, y1 = 0, 1, 1, 0
    while a != 0:
        (q, a), b = divmod(b, a), a
        y0, y1 = y1, y0 - q * y1
        x0, x1 = x1, x0 - q * x1
    return b, x0, y0


def lcm(a, b):
    d = gcd(a, b)
    return a * b // d


def is_even(x):
    return x % 2 == 0


def is_odd(x):
    return x % 2 == 1


def modinv(a, m):
    """return x such that (a * x) % m == 1"""
    g, x, _ = xgcd(a, m)
    if g != 1:
        raise Exception('gcd(a, m) != 1')
    return x % m


def mod_add(x, y):
    x += y
    while x >= MOD:
        x -= MOD
    while x < 0:
        x += MOD
    return x


def mod_mul(x, y):
    return (x * y) % MOD


def mod_pow(x, y):
    if y == 0:
        return 1
    if y % 2:
        return mod_mul(x, mod_pow(x, y - 1))
    p = mod_pow(x, y // 2)
    return mod_mul(p, p)


def mod_inv(y):
    return mod_pow(y, MOD - 2)


def mod_div(x, y):
    # y^(-1): Fermat little theorem, MOD is a prime
    return mod_mul(x, mod_inv(y))

#---------------------------------PRIME---------------------------------------

def is_prime(n):
    if n == 1:
        return False
    for i in range(2, int(n ** 0.5) + 1):
        if n % i:
            return False
    return True


def gen_primes(n):
    """
    generate primes of [1..n] using sieve's method
    """
    P = [True for _ in range(n + 1)]
    P[0] = P[1] = False
    for i in range(int(n ** 0.5) + 1):
        if P[i]:
            for j in range(2 * i, n + 1, i):
                P[j] = False
    return P

#---------------------------------MAIN---------------------------------------

main()