#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % ((int)(1e9) + 7);
    a = (a * a) % ((int)(1e9) + 7);
    b /= 2;
  }
  return res;
}
int n, q;
int arr[123456];
long long newarr[123456];
int leftmost[123456], rightmost[123456];
long long number(int x, int y, int i) {
  int len = y - i + 1;
  return 1LL * len * (i - x + 1);
}
long long solve(int l, int r) {
  for (int i = 0; i < r - l; i++) newarr[i] = abs(arr[i + l] - arr[i + l + 1]);
  stack<pair<long long int, int> > stck, stckroll;
  for (int i = 0; i < r - l; i++) {
    while (stck.empty() == false && stck.top().first < newarr[i]) stck.pop();
    if (stck.empty() == true)
      leftmost[i] = 0;
    else
      leftmost[i] = stck.top().second + 1;
    stck.push(make_pair(newarr[i], i));
  }
  for (int i = r - l - 1; i >= 0; i--) {
    while (stckroll.empty() == false && stckroll.top().first <= newarr[i])
      stckroll.pop();
    if (stckroll.empty() == true)
      rightmost[i] = r - l - 1;
    else
      rightmost[i] = stckroll.top().second - 1;
    stckroll.push(make_pair(newarr[i], i));
  }
  long long ans = 0;
  for (int i = 0; i < r - l; i++) {
    ans += 1LL * number(leftmost[i], rightmost[i], i) * newarr[i];
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    cout << solve(l, r) << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % ((int)(1e9) + 7);
    a = (a * a) % ((int)(1e9) + 7);
    b /= 2;
  }
  return res;
}
int n, q;
int arr[123456];
long long newarr[123456];
int leftmost[123456], rightmost[123456];
long long number(int x, int y, int i) {
  int len = y - i + 1;
  return 1LL * len * (i - x + 1);
}
long long solve(int l, int r) {
  for (int i = 0; i < r - l; i++) newarr[i] = abs(arr[i + l] - arr[i + l + 1]);
  stack<pair<long long int, int> > stck, stckroll;
  for (int i = 0; i < r - l; i++) {
    while (stck.empty() == false && stck.top().first < newarr[i]) stck.pop();
    if (stck.empty() == true)
      leftmost[i] = 0;
    else
      leftmost[i] = stck.top().second + 1;
    stck.push(make_pair(newarr[i], i));
  }
  for (int i = r - l - 1; i >= 0; i--) {
    while (stckroll.empty() == false && stckroll.top().first <= newarr[i])
      stckroll.pop();
    if (stckroll.empty() == true)
      rightmost[i] = r - l - 1;
    else
      rightmost[i] = stckroll.top().second - 1;
    stckroll.push(make_pair(newarr[i], i));
  }
  long long ans = 0;
  for (int i = 0; i < r - l; i++) {
    ans += 1LL * number(leftmost[i], rightmost[i], i) * newarr[i];
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    cout << solve(l, r) << "\n";
  }
  return 0;
}

```