#include <bits/stdc++.h>
using namespace std;
vector<int> tab, lewo, prawo;
vector<long long> con;
vector<pair<int, int> > rob;
int n;
int main() {
  int q, l, r;
  scanf("%d %d", &n, &q);
  tab.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &tab[i]);
  con.resize(n);
  for (int i = 1; i < n; i++) con[i] = abs(tab[i] - tab[i - 1]);
  lewo.resize(n, 0);
  prawo.resize(n, 0);
  for (int i = 1; i < n; i++) {
    while ((rob.size() > 0) && (rob.back().first <= con[i])) rob.pop_back();
    if (rob.size() == 0)
      lewo[i] = 0;
    else
      lewo[i] = rob.back().second;
    rob.push_back(make_pair(con[i], i));
  }
  rob.clear();
  for (int i = n - 1; i >= 0; i--) {
    while ((rob.size() > 0) && (rob.back().first < con[i])) rob.pop_back();
    if (rob.size() == 0)
      prawo[i] = n;
    else
      prawo[i] = rob.back().second;
    rob.push_back(make_pair(con[i], i));
  }
  long long sum = 0;
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &l, &r);
    sum = 0;
    for (int i = l; i < r; i++) {
      sum += con[i] * (i - max(lewo[i], l - 1)) * (min(prawo[i], r) - i);
    }
    cout << sum << "\n";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
vector<int> tab, lewo, prawo;
vector<long long> con;
vector<pair<int, int> > rob;
int n;
int main() {
  int q, l, r;
  scanf("%d %d", &n, &q);
  tab.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &tab[i]);
  con.resize(n);
  for (int i = 1; i < n; i++) con[i] = abs(tab[i] - tab[i - 1]);
  lewo.resize(n, 0);
  prawo.resize(n, 0);
  for (int i = 1; i < n; i++) {
    while ((rob.size() > 0) && (rob.back().first <= con[i])) rob.pop_back();
    if (rob.size() == 0)
      lewo[i] = 0;
    else
      lewo[i] = rob.back().second;
    rob.push_back(make_pair(con[i], i));
  }
  rob.clear();
  for (int i = n - 1; i >= 0; i--) {
    while ((rob.size() > 0) && (rob.back().first < con[i])) rob.pop_back();
    if (rob.size() == 0)
      prawo[i] = n;
    else
      prawo[i] = rob.back().second;
    rob.push_back(make_pair(con[i], i));
  }
  long long sum = 0;
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &l, &r);
    sum = 0;
    for (int i = l; i < r; i++) {
      sum += con[i] * (i - max(lewo[i], l - 1)) * (min(prawo[i], r) - i);
    }
    cout << sum << "\n";
  }
}

```