#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1, L = 18;
long long a[N], d[N];
long long p[N][L];
void build(int len) {
  for (int i = 0; i < len; i++) p[i][0] = i;
  for (int j = 1; (1 << j) - 1 < len; j++)
    for (int i = 0; i + (1 << j) - 1 < len; i++) {
      p[i][j] = p[i][j - 1];
      if (d[p[i][j - 1]] < d[p[i + (1 << (j - 1))][j - 1]])
        p[i][j] = p[i + (1 << (j - 1))][j - 1];
    }
}
int query(int l, int r) {
  int lg = 0;
  while (1 << (lg + 1) <= r - l + 1) lg++;
  int ind1 = p[l][lg], ind2 = p[r - (1 << lg) + 1][lg];
  if (d[ind1] > d[ind2]) return ind1;
  return ind2;
}
long long get(int l, int r) {
  if (l >= r) return 0;
  int ind = query(l, r - 1);
  long long ans =
      1LL * (ind - l + 1) * (r - ind) * d[ind] + get(l, ind) + get(ind + 1, r);
  return ans;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", a + i);
    if (i) d[i - 1] = a[i] - a[i - 1];
    if (d[i - 1] < 0) d[i - 1] *= -1;
  }
  build(n - 1);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%I64d\n", get(--l, --r));
  }
}

### Prompt

Your challenge is to write a Cpp solution to 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;
const int N = 1e5 + 1, L = 18;
long long a[N], d[N];
long long p[N][L];
void build(int len) {
  for (int i = 0; i < len; i++) p[i][0] = i;
  for (int j = 1; (1 << j) - 1 < len; j++)
    for (int i = 0; i + (1 << j) - 1 < len; i++) {
      p[i][j] = p[i][j - 1];
      if (d[p[i][j - 1]] < d[p[i + (1 << (j - 1))][j - 1]])
        p[i][j] = p[i + (1 << (j - 1))][j - 1];
    }
}
int query(int l, int r) {
  int lg = 0;
  while (1 << (lg + 1) <= r - l + 1) lg++;
  int ind1 = p[l][lg], ind2 = p[r - (1 << lg) + 1][lg];
  if (d[ind1] > d[ind2]) return ind1;
  return ind2;
}
long long get(int l, int r) {
  if (l >= r) return 0;
  int ind = query(l, r - 1);
  long long ans =
      1LL * (ind - l + 1) * (r - ind) * d[ind] + get(l, ind) + get(ind + 1, r);
  return ans;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", a + i);
    if (i) d[i - 1] = a[i] - a[i - 1];
    if (d[i - 1] < 0) d[i - 1] *= -1;
  }
  build(n - 1);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%I64d\n", get(--l, --r));
  }
}

```