#include <bits/stdc++.h>
using namespace std;
int n;
long long a[100005], b[100005], q[100005];
long long lft[100005], rgt[100005];
int r, m, k;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  for (int i = 1; i < n; ++i) b[i] = abs(a[i] - a[i + 1]);
  m = n - 1;
  lft[1] = 0;
  r = 1;
  q[1] = 1;
  for (int i = 2; i <= m; ++i) {
    while (r >= 1 && b[i] > b[q[r]]) r--;
    if (r == 0)
      lft[i] = 0;
    else
      lft[i] = q[r];
    ++r;
    q[r] = i;
  }
  rgt[m] = m + 1;
  r = 1;
  q[1] = m;
  for (int i = m - 1; i >= 1; --i) {
    while (r >= 1 && b[i] >= b[q[r]]) r--;
    if (r == 0)
      rgt[i] = m + 1;
    else
      rgt[i] = q[r];
    ++r;
    q[r] = i;
  }
  while (k--) {
    long long l, r;
    long long ans, dql, dqr, dq;
    scanf("%I64d%I64d", &l, &r);
    ans = 0;
    for (int i = l; i < r; ++i) {
      dql = max(lft[i] + 1, l);
      if (rgt[i] == r)
        dqr = r;
      else
        dqr = min(rgt[i], r);
      dq = (i - dql + 1) * (dqr - i);
      dq = dq * b[i];
      ans += dq;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int n;
long long a[100005], b[100005], q[100005];
long long lft[100005], rgt[100005];
int r, m, k;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  for (int i = 1; i < n; ++i) b[i] = abs(a[i] - a[i + 1]);
  m = n - 1;
  lft[1] = 0;
  r = 1;
  q[1] = 1;
  for (int i = 2; i <= m; ++i) {
    while (r >= 1 && b[i] > b[q[r]]) r--;
    if (r == 0)
      lft[i] = 0;
    else
      lft[i] = q[r];
    ++r;
    q[r] = i;
  }
  rgt[m] = m + 1;
  r = 1;
  q[1] = m;
  for (int i = m - 1; i >= 1; --i) {
    while (r >= 1 && b[i] >= b[q[r]]) r--;
    if (r == 0)
      rgt[i] = m + 1;
    else
      rgt[i] = q[r];
    ++r;
    q[r] = i;
  }
  while (k--) {
    long long l, r;
    long long ans, dql, dqr, dq;
    scanf("%I64d%I64d", &l, &r);
    ans = 0;
    for (int i = l; i < r; ++i) {
      dql = max(lft[i] + 1, l);
      if (rgt[i] == r)
        dqr = r;
      else
        dqr = min(rgt[i], r);
      dq = (i - dql + 1) * (dqr - i);
      dq = dq * b[i];
      ans += dq;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```