Problem: There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another. 

Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer. 

When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.

All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.

Input

First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.

Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.

Output

Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.

Example

Input


7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2


Output


55

Note

Let's look at queries in details: 

  1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50. 
  2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48. 
  3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52. 
  4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40. 
  5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40. 
  6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 2e5 + 107;
const double pi = acos(-1.0);
const double eps = 0.0000000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void pr2(T x, int k = 64) {
  long long i;
  for (i = 0; i < k; i++) fprintf(stderr, "%d", (x >> i) & 1);
  putchar(' ');
}
template <typename T>
inline void add_(T &A, int B, long long MOD = M) {
  A += B;
  (A >= MOD) && (A -= MOD);
}
template <typename T>
inline void mul_(T &A, long long B, long long MOD = M) {
  A = (A * B) % MOD;
}
template <typename T>
inline void mod_(T &A, long long MOD = M) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline void max_(T &A, T B) {
  (A < B) && (A = B);
}
template <typename T>
inline void min_(T &A, T B) {
  (A > B) && (A = B);
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = (long long)a * a % M)
    if (b & 1) ret = (long long)ret * a % M;
  return ret;
}
int startTime;
void startTimer() { startTime = clock(); }
void printTimer() {
  fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
}
int n, m, q;
int A[407];
int f[401][401][401];
int main() {
  int i, j, k, l;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= n; j++) f[i][j][0] = A[j] - A[i];
    for (k = 1; k <= n; k++) {
      l = i;
      for (j = i + 1; j <= n; j++) {
        while (l < j && max(f[i][l][k - 1], A[j] - A[l]) >
                            max(f[i][l + 1][k - 1], A[j] - A[l + 1]))
          l++;
        f[i][j][k] = max(f[i][l][k - 1], A[j] - A[l]);
      }
    }
  }
  long long ans = 0;
  for (i = 1; i <= m; i++) {
    int s, f, c, r;
    scanf("%d%d%d%d", &s, &f, &c, &r);
    ans = max(ans, (long long)::f[s][f][r] * c);
  }
  printf("%I64d\n", ans);
}