#include <bits/stdc++.h>
using namespace std;
long long a[100010], b[100010];
long long l[100010], r[100010];
stack<int> s;
int main() {
  int n, q, i, j, x, y;
  scanf("%d", &n);
  scanf("%d", &q);
  for (i = 0; i < n; ++i) scanf("%lld", &b[i]);
  for (i = 0; i < n - 1; ++i) a[i] = abs(b[i + 1] - b[i]);
  for (i = 0; i < n - 1; ++i) {
    while (!s.empty()) {
      if (a[s.top()] <= a[i]) {
        r[s.top()] = i - 1;
        s.pop();
      } else {
        l[i] = s.top() + 1;
        break;
      }
    }
    if (s.empty()) l[i] = 0;
    s.push(i);
  }
  while (!s.empty()) {
    r[s.top()] = n - 2;
    s.pop();
  }
  while (q--) {
    scanf("%d", &x);
    scanf("%d", &y);
    --x;
    --y;
    --y;
    long long ans = 0;
    for (i = x; i <= y; ++i) {
      long long v1 = i - max(l[i], (long long)x) + 1;
      long long v2 = min(r[i], (long long)y) - i + 1;
      ans = ans + v1 * v2 * a[i];
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Please formulate 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;
long long a[100010], b[100010];
long long l[100010], r[100010];
stack<int> s;
int main() {
  int n, q, i, j, x, y;
  scanf("%d", &n);
  scanf("%d", &q);
  for (i = 0; i < n; ++i) scanf("%lld", &b[i]);
  for (i = 0; i < n - 1; ++i) a[i] = abs(b[i + 1] - b[i]);
  for (i = 0; i < n - 1; ++i) {
    while (!s.empty()) {
      if (a[s.top()] <= a[i]) {
        r[s.top()] = i - 1;
        s.pop();
      } else {
        l[i] = s.top() + 1;
        break;
      }
    }
    if (s.empty()) l[i] = 0;
    s.push(i);
  }
  while (!s.empty()) {
    r[s.top()] = n - 2;
    s.pop();
  }
  while (q--) {
    scanf("%d", &x);
    scanf("%d", &y);
    --x;
    --y;
    --y;
    long long ans = 0;
    for (i = x; i <= y; ++i) {
      long long v1 = i - max(l[i], (long long)x) + 1;
      long long v2 = min(r[i], (long long)y) - i + 1;
      ans = ans + v1 * v2 * a[i];
    }
    printf("%lld\n", ans);
  }
}

```