#include <bits/stdc++.h>
using namespace std;
long long n, q, ans, l, r, a[100005], diff[100005], low_bound[100005],
    high_bound[100005], high, low;
stack<long long> sta, sta2;
int main() {
  scanf("%lld", &n), scanf("%lld", &q);
  for (long long i = 1; i < n + 1; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i < n; i++) diff[i] = abs(a[i + 1] - a[i]);
  sta.push(0), sta2.push(n);
  for (long long i = 1; i < n; i++) {
    while (sta.top() >= 1 && diff[i] >= diff[sta.top()]) sta.pop();
    high_bound[i] = sta.top(), sta.push(i);
  }
  for (long long i = n - 1; i >= 1; i--) {
    while (sta2.top() < n && diff[i] > diff[sta2.top()]) sta2.pop();
    low_bound[i] = sta2.top(), sta2.push(i);
  }
  while (q--) {
    scanf("%lld", &l), scanf("%lld", &r), --r, ans = 0;
    for (long long i = l; i < r + 1; i++) {
      if (low_bound[i] <= (r + 1))
        high = low_bound[i] - 1;
      else
        high = r;
      if (high_bound[i] >= (l - 1))
        low = high_bound[i] + 1;
      else
        low = l;
      ans += (diff[i] * (high - i + 1) * (i - low + 1));
    }
    printf("%lld ", ans), printf("\n");
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 n, q, ans, l, r, a[100005], diff[100005], low_bound[100005],
    high_bound[100005], high, low;
stack<long long> sta, sta2;
int main() {
  scanf("%lld", &n), scanf("%lld", &q);
  for (long long i = 1; i < n + 1; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i < n; i++) diff[i] = abs(a[i + 1] - a[i]);
  sta.push(0), sta2.push(n);
  for (long long i = 1; i < n; i++) {
    while (sta.top() >= 1 && diff[i] >= diff[sta.top()]) sta.pop();
    high_bound[i] = sta.top(), sta.push(i);
  }
  for (long long i = n - 1; i >= 1; i--) {
    while (sta2.top() < n && diff[i] > diff[sta2.top()]) sta2.pop();
    low_bound[i] = sta2.top(), sta2.push(i);
  }
  while (q--) {
    scanf("%lld", &l), scanf("%lld", &r), --r, ans = 0;
    for (long long i = l; i < r + 1; i++) {
      if (low_bound[i] <= (r + 1))
        high = low_bound[i] - 1;
      else
        high = r;
      if (high_bound[i] >= (l - 1))
        low = high_bound[i] + 1;
      else
        low = l;
      ans += (diff[i] * (high - i + 1) * (i - low + 1));
    }
    printf("%lld ", ans), printf("\n");
  }
  return 0;
}

```