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;
using ll = long long;
using ld = long double;
using pii = pair<ll, ll>;
using pll = pair<ll, ll>;
using pli = pair<ll, ll>;
using vi = vector<ll>;
using vl = vector<ll>;
using vpii = vector<pii>;
template <class T>
inline bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool ckmax(T& a, const T& b) {
  return b > a ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const char nl = '\n';
const ll mxN = 401;
const ll MOD = 1e9 + 7;
const ll infLL = 1e18;
const ld eps = 1e-6;
ll n, m;
ll a[mxN];
vector<array<ll, 3>> v[mxN];
ll dp[mxN][mxN];
ll ans = 0;
ll calc(ll p, ll c, ll q) { return max(dp[q - 1][p], (ll)a[c] - a[p]); }
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> m;
  for (ll i = (0); i < (n); i++) {
    cin >> a[i];
  }
  for (ll i = (0); i < (m); i++) {
    ll S, F, C, R;
    cin >> S >> F >> C >> R;
    ckmin(R, n - 1);
    S--;
    F--;
    v[S].push_back({F, C, R});
  }
  for (ll i = (0); i < (n); i++) {
    for (ll l = (i + 1); l < (n); l++) {
      dp[0][l] = a[l] - a[i];
    }
    for (ll k = (1); k < (n); k++) {
      ll op = i;
      for (ll r = (i + 1); r < (n); r++) {
        while (op < r && calc(op, r, k) >= calc(op + 1, r, k)) {
          op++;
        }
        dp[k][r] = calc(op, r, k);
      }
    }
    for (auto [F, C, R] : v[i]) {
      ckmax(ans, 1LL * C * dp[R][F]);
    }
  }
  cout << ans << nl;
}