#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int a[MAXN], b[MAXN], p[MAXN], f[MAXN];
int main() {
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) b[i] = abs(a[i] - a[i - 1]);
  vector<pair<int, int> > v;
  for (int i = 1; i < n; i++) {
    while (v.size() and v.back().first <= b[i]) v.pop_back();
    if (v.empty())
      p[i] = 1;
    else
      p[i] = v.back().second + 1;
    v.push_back({b[i], i});
  }
  v.clear();
  for (int i = n - 1; i > 0; i--) {
    while (v.size() and v.back().first < b[i]) v.pop_back();
    if (v.empty())
      f[i] = n - 1;
    else
      f[i] = v.back().second - 1;
    v.push_back({b[i], i});
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    long long ans = 0;
    cin >> l >> r;
    l--;
    r--;
    for (int j = l + 1; j <= r; j++) {
      int s = max(l + 1, p[j]), e = min(r, f[j]);
      if (s > e) continue;
      long long x = j - s, y = e - j;
      ans += (x * y + x + y + 1ll) * b[j];
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const int MAXN = 1e5 + 10;
int a[MAXN], b[MAXN], p[MAXN], f[MAXN];
int main() {
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) b[i] = abs(a[i] - a[i - 1]);
  vector<pair<int, int> > v;
  for (int i = 1; i < n; i++) {
    while (v.size() and v.back().first <= b[i]) v.pop_back();
    if (v.empty())
      p[i] = 1;
    else
      p[i] = v.back().second + 1;
    v.push_back({b[i], i});
  }
  v.clear();
  for (int i = n - 1; i > 0; i--) {
    while (v.size() and v.back().first < b[i]) v.pop_back();
    if (v.empty())
      f[i] = n - 1;
    else
      f[i] = v.back().second - 1;
    v.push_back({b[i], i});
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    long long ans = 0;
    cin >> l >> r;
    l--;
    r--;
    for (int j = l + 1; j <= r; j++) {
      int s = max(l + 1, p[j]), e = min(r, f[j]);
      if (s > e) continue;
      long long x = j - s, y = e - j;
      ans += (x * y + x + y + 1ll) * b[j];
    }
    cout << ans << endl;
  }
  return 0;
}

```