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 NMAX = 400;
const int QMAX = 25e4;
const int LEN = 1e4;
const long long inf = 1LL << 60;
char buff[LEN];
int ind = LEN - 1;
int i32() {
  int ans = 0;
  while (buff[ind] < '0' || buff[ind] > '9') {
    if (++ind >= LEN) {
      ind = 0;
      fread(buff, 1, LEN, stdin);
    }
  }
  while (!(buff[ind] < '0' || buff[ind] > '9')) {
    ans = ans * 10 + (buff[ind] - '0');
    if (++ind >= LEN) {
      ind = 0;
      fread(buff, 1, LEN, stdin);
    }
  }
  return ans;
}
int n, q;
int a[NMAX + 5];
long long dp[2][NMAX + 5][NMAX + 5];
long long pref_sum[NMAX + 5];
vector<pair<pair<int, int>, int> > v[NMAX + 5];
int main() {
  n = i32();
  q = i32();
  for (int i = 1; i <= n; i++) {
    a[i] = i32();
    pref_sum[i] = a[i];
  }
  for (int j = 1; j <= n; j++) {
    for (int k = 1; k <= j; k++) {
      dp[1][j][k] = inf;
    }
    for (int k = j + 1; k <= n; k++) {
      dp[1][j][k] = a[k] - a[j];
    }
  }
  long long ans = 0;
  while (q--) {
    int s, f, c, r;
    s = i32();
    f = i32();
    c = i32();
    r = i32();
    v[min(f - s, r + 1)].push_back({{s, f}, c});
  }
  for (int i = 2, l = 0; i <= n; i++, l ^= 1) {
    for (int j = 1; j <= n; j++) {
      int pref = j;
      for (int k = 1; k <= j; k++) {
        dp[l][j][k] = inf;
      }
      for (int k = j + 1; k <= n; k++) {
        while (pref + 1 < k &&
               dp[l ^ 1][j][pref + 1] <= pref_sum[k] - pref_sum[pref + 1]) {
          pref++;
        }
        dp[l][j][k] = min(pref_sum[k] - pref_sum[pref], dp[l ^ 1][j][pref + 1]);
        dp[l][j][k] = min(dp[l][j][k], dp[l ^ 1][j][k]);
      }
    }
    for (auto it : v[i]) {
      ans = max(ans, 1LL * dp[l][it.first.first][it.first.second] * it.second);
    }
  }
  for (auto it : v[1]) {
    ans = max(ans, 1LL * (a[it.first.second] - a[it.first.first]) * it.second);
  }
  printf("%lld\n", ans);
  return 0;
}