#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const int N = 1e5 + 10, INFI = 1e9 + 10;
int n, q, a[N], l[N], r[N], s[N];
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    s[i + 1] = abs(a[i] - a[i + 1]);
  }
  s[0] = s[n] = INFI;
  for (int i = n - 1; i >= 0; i--) {
    r[i] = i + 1;
    while (r[i] < n && s[i] >= s[r[i]]) {
      r[i] = r[r[i]];
    }
  }
  for (int i = 1; i < n; i++) {
    l[i] = i - 1;
    while (l[i] > 0 && s[i] > s[l[i]]) {
      l[i] = l[l[i]];
    }
  }
  for (int i = 0; i < q; i++) {
    long long ans = 0;
    int b, e;
    cin >> b >> e;
    for (int i = b; i < e; i++) {
      int rsz = min(r[i] - 1, e - 1) - i;
      int lsz = i - max(l[i] + 1, b);
      ans += 1LL * (1LL * lsz * rsz + 1LL * lsz + 1LL * rsz + 1) * s[i];
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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 long long INF = 1e18;
const int N = 1e5 + 10, INFI = 1e9 + 10;
int n, q, a[N], l[N], r[N], s[N];
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    s[i + 1] = abs(a[i] - a[i + 1]);
  }
  s[0] = s[n] = INFI;
  for (int i = n - 1; i >= 0; i--) {
    r[i] = i + 1;
    while (r[i] < n && s[i] >= s[r[i]]) {
      r[i] = r[r[i]];
    }
  }
  for (int i = 1; i < n; i++) {
    l[i] = i - 1;
    while (l[i] > 0 && s[i] > s[l[i]]) {
      l[i] = l[l[i]];
    }
  }
  for (int i = 0; i < q; i++) {
    long long ans = 0;
    int b, e;
    cin >> b >> e;
    for (int i = b; i < e; i++) {
      int rsz = min(r[i] - 1, e - 1) - i;
      int lsz = i - max(l[i] + 1, b);
      ans += 1LL * (1LL * lsz * rsz + 1LL * lsz + 1LL * rsz + 1) * s[i];
    }
    cout << ans << '\n';
  }
  return 0;
}

```