#include <bits/stdc++.h>
using namespace std;
int n, m, ta, tb, y[100005], z[100005];
long long x[100005], ans;
vector<int> la;
int main() {
  scanf("%d%d", &n, &m);
  for (int a = 0; a < n; a++) scanf("%I64d", &x[a]);
  n--;
  for (int a = 0; a < n; a++) x[a] = abs(x[a] - x[a + 1]);
  la.clear();
  for (int a = 0; a < n; a++) {
    while (!la.empty() && x[a] > x[la.back()]) {
      y[la.back()] = a - 1;
      la.pop_back();
    }
    la.push_back(a);
  }
  while (!la.empty()) {
    y[la.back()] = n - 1;
    la.pop_back();
  }
  for (int a = n - 1; a >= 0; a--) {
    while (!la.empty() && x[a] >= x[la.back()]) {
      z[la.back()] = a + 1;
      la.pop_back();
    }
    la.push_back(a);
  }
  while (!la.empty()) {
    z[la.back()] = 0;
    la.pop_back();
  }
  while (m--) {
    scanf("%d%d", &ta, &tb);
    ans = 0;
    ta--;
    tb -= 2;
    for (int a = ta; a <= tb; a++)
      ans +=
          x[a] * min(a - ta + 1, a - z[a] + 1) * min(tb - a + 1, y[a] - a + 1);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int n, m, ta, tb, y[100005], z[100005];
long long x[100005], ans;
vector<int> la;
int main() {
  scanf("%d%d", &n, &m);
  for (int a = 0; a < n; a++) scanf("%I64d", &x[a]);
  n--;
  for (int a = 0; a < n; a++) x[a] = abs(x[a] - x[a + 1]);
  la.clear();
  for (int a = 0; a < n; a++) {
    while (!la.empty() && x[a] > x[la.back()]) {
      y[la.back()] = a - 1;
      la.pop_back();
    }
    la.push_back(a);
  }
  while (!la.empty()) {
    y[la.back()] = n - 1;
    la.pop_back();
  }
  for (int a = n - 1; a >= 0; a--) {
    while (!la.empty() && x[a] >= x[la.back()]) {
      z[la.back()] = a + 1;
      la.pop_back();
    }
    la.push_back(a);
  }
  while (!la.empty()) {
    z[la.back()] = 0;
    la.pop_back();
  }
  while (m--) {
    scanf("%d%d", &ta, &tb);
    ans = 0;
    ta--;
    tb -= 2;
    for (int a = ta; a <= tb; a++)
      ans +=
          x[a] * min(a - ta + 1, a - z[a] + 1) * min(tb - a + 1, y[a] - a + 1);
    printf("%I64d\n", ans);
  }
  return 0;
}

```