#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    vector<pii> s;
    ll t = 0;
    s.push_back(pii{1e9, l});
    ll res = 0;
    for (int i = l + 1; i <= r; ++i) {
      int cur = abs(a[i] - a[i - 1]);
      pii last = s.back();
      while (last.first <= cur) {
        pii prev = s[s.size() - 2];
        t -= 1LL * last.first * (last.second - prev.second);
        s.resize(s.size() - 1);
        last = s.back();
      }
      t += 1LL * cur * (i - last.second);
      s.push_back(pii{cur, i});
      res += t;
    }
    cout << res << '\n';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

A function <image> is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all <image>. We'll deal with a more... discrete version of this term.

For an array <image>, we define it's Lipschitz constant <image> as follows:

  * if n < 2, <image>
  * if n ≥ 2, <image> over all 1 ≤ i < j ≤ n



In other words, <image> is the smallest non-negative integer such that |h[i] - h[j]| ≤ L·|i - j| holds for all 1 ≤ i, j ≤ n.

You are given an array <image> of size n and q queries of the form [l, r]. For each query, consider the subarray <image>; determine the sum of Lipschitz constants of all subarrays of <image>.

Input

The first line of the input contains two space-separated integers n and q (2 ≤ n ≤ 100 000 and 1 ≤ q ≤ 100) — the number of elements in array <image> and the number of queries respectively.

The second line contains n space-separated integers <image> (<image>).

The following q lines describe queries. The i-th of those lines contains two space-separated integers li and ri (1 ≤ li < ri ≤ n).

Output

Print the answers to all queries in the order in which they are given in the input. For the i-th query, print one line containing a single integer — the sum of Lipschitz constants of all subarrays of <image>.

Examples

Input

10 4
1 5 2 9 1 3 4 2 1 7
2 4
3 8
7 10
1 9


Output

17
82
23
210


Input

7 6
5 7 7 4 6 6 2
1 2
2 3
2 6
1 7
4 7
3 5


Output

2
0
22
59
16
8

Note

In the first query of the first sample, the Lipschitz constants of subarrays of <image> with length at least 2 are:

  * <image>
  * <image>
  * <image>



The answer to the query is their sum.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    vector<pii> s;
    ll t = 0;
    s.push_back(pii{1e9, l});
    ll res = 0;
    for (int i = l + 1; i <= r; ++i) {
      int cur = abs(a[i] - a[i - 1]);
      pii last = s.back();
      while (last.first <= cur) {
        pii prev = s[s.size() - 2];
        t -= 1LL * last.first * (last.second - prev.second);
        s.resize(s.size() - 1);
        last = s.back();
      }
      t += 1LL * cur * (i - last.second);
      s.push_back(pii{cur, i});
      res += t;
    }
    cout << res << '\n';
  }
  return 0;
}

```