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;
inline void readI(int *i) {
  register int t = 0;
  register char z = getchar();
  int znak = 1;
  if (z == '-') {
    znak = -1;
  }
  while ((z < '0') || ('9' < z)) {
    z = getchar();
    if (z == '-') {
      znak = -1;
    }
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *i = (t * znak);
}
inline void readUI(int *i) {
  register int t = 0;
  register char z = getchar();
  while ((z < '0') || ('9' < z)) {
    z = getchar();
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *i = t;
}
inline void readL(long long int *l) {
  register long long int t = 0;
  register char z = getchar();
  int znak = 1;
  if (z == '-') {
    znak *= -1;
  }
  while ((z < '0') || ('9' < z)) {
    z = getchar();
    if (z == '-') {
      znak = -1;
    }
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *l = (t * znak);
}
inline void readUL(long long int *l) {
  register long long int t = 0;
  register char z = getchar();
  while ((z < '0') || ('9' < z)) {
    z = getchar();
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *l = t;
}
inline void writeL(long long int l) {
  if (l == 0) {
    putchar(48);
  } else {
    if (l < 0) {
      l *= -1;
      putchar(45);
    }
    int _tab[21];
    int wsk = 0;
    while (l > 0) {
      ++wsk;
      _tab[wsk] = (l % 10) + 48;
      l /= 10;
    }
    for (int j = wsk; j >= 1; --j) {
      putchar(_tab[j]);
    }
  }
}
inline void writeS(string s) {
  int l = s.length();
  for (int i = 0; i < l; ++i) {
    putchar(s[i]);
  }
}
inline void space() { putchar(32); }
inline void endl() { putchar(10); }
int n, m, s, f, c, koszt;
int tab[405];
int dp[405][405][2];
long long int odp;
vector<pair<pair<int, int>, int> > v[405];
int query(int s, int f) {
  int ret = 0;
  for (int i = s + 1; i <= f; ++i) {
    ret = max(ret, tab[i] - tab[i - 1]);
  }
  return ret;
}
int main() {
  readUI(&n);
  readUI(&m);
  for (int i = 1; i <= n; ++i) {
    readUI(&tab[i]);
  }
  for (int i = 1; i <= m; ++i) {
    readUI(&s);
    readUI(&f);
    readUI(&koszt);
    readUI(&c);
    ++c;
    if (c >= f - s) {
      odp = max(odp, (long long int)query(s, f) * koszt);
    } else if (c == 1) {
      odp = max(odp, (long long int)(tab[f] - tab[s]) * koszt);
    } else {
      v[c].emplace_back(make_pair(make_pair(s, f), koszt));
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      dp[i][j][1] = tab[j] - tab[i];
    }
  }
  for (int r = 2; r < n; ++r) {
    sort(v[r].begin(), v[r].end());
    int l = v[r].size();
    int poz = 0;
    if (0) {
      fprintf(stderr, "----------------r=%d l=%d\n", r, l);
    }
    for (int a = 1; a <= n - r; ++a) {
      int wsk = a + r - 1;
      for (int b = a + r; b <= n; ++b) {
        dp[a][b][r & 1] = max(dp[a][wsk][r & 1 ^ 1], tab[b] - tab[wsk]);
        while (wsk < b - 1 && max(dp[a][wsk + 1][r & 1 ^ 1],
                                  tab[b] - tab[wsk + 1]) < dp[a][b][r & 1]) {
          dp[a][b][r & 1] =
              max(dp[a][wsk + 1][r & 1 ^ 1], tab[b] - tab[wsk + 1]);
          ++wsk;
        }
        if (0) {
          fprintf(stderr, "a=%d b=%d r=%d dp[a][b][r]=%d\n", a, b, r,
                  dp[a][b][r & 1]);
        }
        if (0)
          if (poz < l) {
            fprintf(stderr, "poz=%d v[r][poz]=(%d, %d, %d)\n", poz,
                    v[r][poz].first.first, v[r][poz].first.second,
                    v[r][poz].second);
          }
        if (poz < l) {
          if (a == v[r][poz].first.first && b == v[r][poz].first.second) {
            odp = max(odp, (long long int)dp[a][b][r & 1] * v[r][poz].second);
            ++poz;
          }
        }
      }
    }
  }
  writeL(odp);
  return 0;
}