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>
template <class T>
inline void read(T &x) {
  int c, f = 1;
  while (!isdigit(c = getchar())) {
    if (c == '-') f = -1;
  }
  x = c - '0';
  while (isdigit(c = getchar())) {
    x = x * 10 + c - '0';
  }
  x *= f;
}
using LL = long long;
const int max_N = 400 + 21;
const int max_M = (int)3e5 + 21;
int n, m, a[max_N], cur, cur_r, dp[2][max_N][max_N];
LL ans;
int s[max_M], t[max_M], c[max_M], r[max_M], Q[max_M];
struct node {
  int x, y;
  explicit node(int x = 0, int y = 0) : x(x), y(y) {}
  bool operator<(const node &o) const { return x + a[y] < o.x + a[y]; }
} tmp[max_N];
int tmp1[max_N], tmp2;
void solve() {
  ++cur_r;
  cur ^= 1;
  for (int u = 1; u <= n; ++u) {
    for (int v = u; v <= n; ++v) {
      dp[cur][u][v] = dp[cur ^ 1][u][v];
      tmp[v] = node(dp[cur ^ 1][u][v], v);
    }
    std::sort(tmp + u, tmp + n + 1);
    tmp2 = -1;
    tmp1[n] = tmp[n].x;
    for (int v = n - 1; v >= u; --v) {
      tmp1[v] = std::min(tmp1[v + 1], tmp[v].x);
    }
    int i = u;
    for (int v = u; v <= n; ++v) {
      while (i <= n && tmp[i].x <= a[v] - a[tmp[i].y]) {
        tmp2 = std::max(tmp2, a[tmp[i++].y]);
      }
      if (i <= n) {
        dp[cur][u][v] = std::min(dp[cur][u][v], tmp1[i]);
      }
      if (tmp2 != -1) {
        dp[cur][u][v] = std::min(dp[cur][u][v], a[v] - tmp2);
      }
    }
  }
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
  for (int i = 1; i <= m; ++i) {
    read(s[i]), read(t[i]), read(c[i]), read(r[i]);
    Q[i] = i;
  }
  std::sort(Q + 1, Q + 1 + m, [&](int x, int y) { return r[x] < r[y]; });
  for (int u = 1; u <= n; ++u) {
    for (int v = u; v <= n; ++v) {
      dp[cur][u][v] = a[v] - a[u];
    }
  }
  for (int i = 1; i <= m; ++i) {
    while (cur_r < r[Q[i]]) solve();
    int u = s[Q[i]], v = t[Q[i]];
    if (u > v) std::swap(u, v);
    ans = std::max(ans, 1ll * c[Q[i]] * dp[cur][u][v]);
  }
  printf("%lld\n", ans);
}