#include <bits/stdc++.h>
using namespace std;
int n, q, pos[100010];
long long a[100010], b[100010], R[100010], L[100010], stk[100010];
int main() {
  scanf("%d%d", &n, &q);
  for (register int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  while (q--) {
    int l, r;
    memset(R, 0, sizeof(R));
    memset(L, 0, sizeof(L));
    scanf("%d%d", &l, &r);
    r--;
    long long ans = 0;
    for (register int i = l, j = 1; i <= r; i++, j++)
      b[j] = abs(a[i + 1] - a[i]);
    int tot = r - l + 1, tp = 0;
    for (register int i = 1; i <= tot; i++) {
      while (tp && stk[tp] < b[i]) R[pos[tp]] = i - pos[tp], tp--;
      L[i] = i - pos[tp];
      stk[++tp] = b[i];
      pos[tp] = i;
    }
    while (tp) R[pos[tp]] = tot + 1 - pos[tp], tp--;
    for (register int i = 1; i <= tot; i++) ans = ans + (b[i] * R[i] * L[i]);
    printf("%lld\n", ans);
  }
}

### 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;
int n, q, pos[100010];
long long a[100010], b[100010], R[100010], L[100010], stk[100010];
int main() {
  scanf("%d%d", &n, &q);
  for (register int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  while (q--) {
    int l, r;
    memset(R, 0, sizeof(R));
    memset(L, 0, sizeof(L));
    scanf("%d%d", &l, &r);
    r--;
    long long ans = 0;
    for (register int i = l, j = 1; i <= r; i++, j++)
      b[j] = abs(a[i + 1] - a[i]);
    int tot = r - l + 1, tp = 0;
    for (register int i = 1; i <= tot; i++) {
      while (tp && stk[tp] < b[i]) R[pos[tp]] = i - pos[tp], tp--;
      L[i] = i - pos[tp];
      stk[++tp] = b[i];
      pos[tp] = i;
    }
    while (tp) R[pos[tp]] = tot + 1 - pos[tp], tp--;
    for (register int i = 1; i <= tot; i++) ans = ans + (b[i] * R[i] * L[i]);
    printf("%lld\n", ans);
  }
}

```