#include <bits/stdc++.h>
using namespace std;
long long a[100000 + 20];
long long b[100000 + 20];
int rm[100000 + 20];
int lm[100000 + 20];
int main() {
  int n, Q;
  scanf("%d%d", &n, &Q);
  vector<pair<int, int>> W;
  for (int w = 0; w < n; w++) {
    scanf("%d", &a[w]);
  }
  for (int i = 0; i < n - 1; i++) {
    b[i] = abs(a[i + 1] - a[i]);
  }
  memset(lm, 0, sizeof(lm));
  memset(rm, 0, sizeof(rm));
  for (int i = 0; i < n; i++) {
    while (!W.empty() && W.back().second < b[i]) {
      rm[W.back().first] = i;
      W.pop_back();
    }
    W.push_back(make_pair(i, b[i]));
  }
  for (int i = 0; i < W.size(); i++) {
    rm[W[i].first] = n;
  }
  W.clear();
  for (int i = n - 1; i >= 0; i--) {
    while (!W.empty() && W.back().second <= b[i]) {
      lm[W.back().first] = i;
      W.pop_back();
    }
    W.push_back(make_pair(i, b[i]));
  }
  for (int i = 0; i < W.size(); i++) {
    lm[W[i].first] = -1;
  }
  long long anw = 0;
  for (int i = 0; i < Q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    for (int i = l; i < r; i++) {
      anw += (((long long)((b[i]) * ((i - max(lm[i] + 1, l) + 1)))) *
              ((long long)((min(r - 1, rm[i] - 1) - i) + 1)));
    }
    printf("%I64d\n", anw);
    anw = 0;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
long long a[100000 + 20];
long long b[100000 + 20];
int rm[100000 + 20];
int lm[100000 + 20];
int main() {
  int n, Q;
  scanf("%d%d", &n, &Q);
  vector<pair<int, int>> W;
  for (int w = 0; w < n; w++) {
    scanf("%d", &a[w]);
  }
  for (int i = 0; i < n - 1; i++) {
    b[i] = abs(a[i + 1] - a[i]);
  }
  memset(lm, 0, sizeof(lm));
  memset(rm, 0, sizeof(rm));
  for (int i = 0; i < n; i++) {
    while (!W.empty() && W.back().second < b[i]) {
      rm[W.back().first] = i;
      W.pop_back();
    }
    W.push_back(make_pair(i, b[i]));
  }
  for (int i = 0; i < W.size(); i++) {
    rm[W[i].first] = n;
  }
  W.clear();
  for (int i = n - 1; i >= 0; i--) {
    while (!W.empty() && W.back().second <= b[i]) {
      lm[W.back().first] = i;
      W.pop_back();
    }
    W.push_back(make_pair(i, b[i]));
  }
  for (int i = 0; i < W.size(); i++) {
    lm[W[i].first] = -1;
  }
  long long anw = 0;
  for (int i = 0; i < Q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    for (int i = l; i < r; i++) {
      anw += (((long long)((b[i]) * ((i - max(lm[i] + 1, l) + 1)))) *
              ((long long)((min(r - 1, rm[i] - 1) - i) + 1)));
    }
    printf("%I64d\n", anw);
    anw = 0;
  }
  return 0;
}

```