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 N = 4e2 + 10;
const int MOD = 1e9 + 7;
const int base = 257;
const int M = 1e6 + 1;
const long long INFI = 1e18;
long long Hash[M], po[M];
long long gethash(long long l, long long r) {
  return ((Hash[r] - Hash[l - 1] * po[r - l + 1] + MOD * MOD) % MOD);
}
void precomputehashing() {
  po[0] = 1;
  for (int i = 1; i <= M; i++) po[i] = (po[i - 1] * base) % MOD;
  string s;
  s = " " + s;
  for (int i = 1; i <= s.size() - 1; i++)
    Hash[i] = (Hash[i - 1] * base + s[i] - '*' + 1) % MOD;
}
long long fac[N], ifac[N];
long long PowerMod(long long a, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) {
      ret *= a;
      ret %= MOD;
    }
    a *= a;
    a %= MOD;
    n /= 2;
  }
  return ret;
}
void precomputecombinatoris() {
  int i;
  fac[0] = 1;
  for (i = 1; i < N; i++) {
    fac[i] = (i * fac[i - 1]) % MOD;
  }
  ifac[N - 1] = PowerMod(fac[N - 1], MOD - 2);
  for (i = N - 2; i >= 0; i--) {
    ifac[i] = ((i + 1) * ifac[i + 1]) % MOD;
  }
}
long long C(int r, int n) {
  long long ret = fac[n];
  ret *= ifac[r];
  ret %= MOD;
  ret *= ifac[n - r];
  ret %= MOD;
  return ret;
}
struct data {
  int s, f, c, r;
};
bool cmp(data A, data B) { return A.r < B.r; }
data q[M];
long long dp[N][N][2], pos[N][N][2], a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> q[i].s >> q[i].f >> q[i].c >> q[i].r;
  }
  sort(q + 1, q + m + 1, cmp);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      for (int k = 0; k <= 1; k++) {
        dp[i][j][k] = INFI;
        dp[i][i][k] = 0;
      }
    }
  }
  long long res = 0;
  int cur = 1;
  for (int k = 1; k <= q[m].r + 1; k++) {
    for (int cnt = 1; cnt <= n; cnt++) {
      for (int i = 1; i + cnt - 1 <= n; i++) {
        int j = i + cnt - 1;
        int cur = pos[i][j - 1][1];
        while (cur + 1 < j && max(dp[i][cur][0], a[j] - a[cur]) >=
                                  max(dp[i][cur + 1][0], a[j] - a[cur + 1])) {
          cur++;
        }
        pos[i][j][1] = cur;
        dp[i][j][1] = max(dp[i][cur][0], a[j] - a[cur]);
      }
    }
    while (true) {
      if (cur > m || q[cur].r != k - 1) {
        break;
      }
      res = max(res, dp[q[cur].s][q[cur].f][1] * q[cur].c);
      cur++;
    }
    for (int cnt = max(k, 2); cnt <= n; cnt++) {
      for (int i = 1; i + cnt - 1 <= n; i++) {
        int j = i + cnt - 1;
        dp[i][j][0] = dp[i][j][1];
        pos[i][j][0] = pos[i][j][1];
        dp[i][j][1] = INFI;
        pos[i][j][1] = 0;
      }
    }
  }
  cout << res;
  return 0;
}