#include <bits/stdc++.h>
using namespace std;
int n, q;
int a[100007];
int t[100007];
int lewo[100007];
int prawo[100007];
long long query(int l, int r) {
  memset(t, 0, sizeof t);
  int d = r - l + 1;
  for (int i = 1; i < d; ++i) {
    t[i] = abs(a[l + i - 1] - a[l + i]);
  }
  lewo[1] = 0;
  for (int i = 2; i < d; ++i) {
    int j = i - 1;
    while (j > 0 && make_pair(t[i], i) > make_pair(t[j], j)) j = lewo[j];
    lewo[i] = j;
  }
  prawo[d - 1] = d;
  for (int i = d - 2; i >= 1; --i) {
    int j = i + 1;
    while (j < d && make_pair(t[i], i) > make_pair(t[j], j)) j = prawo[j];
    prawo[i] = j;
  }
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    int p = lewo[i] + 1, q = prawo[i];
    if (p != q) res += (long long)t[i] * (i - p + 1) * (q - i);
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << query(l, r) << '\n';
  }
}

### 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;
int n, q;
int a[100007];
int t[100007];
int lewo[100007];
int prawo[100007];
long long query(int l, int r) {
  memset(t, 0, sizeof t);
  int d = r - l + 1;
  for (int i = 1; i < d; ++i) {
    t[i] = abs(a[l + i - 1] - a[l + i]);
  }
  lewo[1] = 0;
  for (int i = 2; i < d; ++i) {
    int j = i - 1;
    while (j > 0 && make_pair(t[i], i) > make_pair(t[j], j)) j = lewo[j];
    lewo[i] = j;
  }
  prawo[d - 1] = d;
  for (int i = d - 2; i >= 1; --i) {
    int j = i + 1;
    while (j < d && make_pair(t[i], i) > make_pair(t[j], j)) j = prawo[j];
    prawo[i] = j;
  }
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    int p = lewo[i] + 1, q = prawo[i];
    if (p != q) res += (long long)t[i] * (i - p + 1) * (q - i);
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << query(l, r) << '\n';
  }
}

```