#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
int n;
int a[N];
long long query(int l, int r) {
  vector<pair<long long, long long> > stack;
  long long ret = 0;
  long long cur = 0;
  for (int i = l + 1; i <= r; ++i) {
    long long diff = abs(a[i] - a[i - 1]);
    long long tot = 1;
    for (; stack.size();) {
      if (stack.back().first <= diff) {
        tot += stack.back().second;
        cur -= stack.back().first * stack.back().second;
        stack.pop_back();
      } else {
        break;
      }
    }
    stack.push_back(make_pair(diff, tot));
    cur += diff * tot;
    ret += cur;
  }
  return ret;
}
void solve() {
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  for (; q--;) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, --r;
    cout << query(l, r) << endl;
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 100000 + 10;
int n;
int a[N];
long long query(int l, int r) {
  vector<pair<long long, long long> > stack;
  long long ret = 0;
  long long cur = 0;
  for (int i = l + 1; i <= r; ++i) {
    long long diff = abs(a[i] - a[i - 1]);
    long long tot = 1;
    for (; stack.size();) {
      if (stack.back().first <= diff) {
        tot += stack.back().second;
        cur -= stack.back().first * stack.back().second;
        stack.pop_back();
      } else {
        break;
      }
    }
    stack.push_back(make_pair(diff, tot));
    cur += diff * tot;
    ret += cur;
  }
  return ret;
}
void solve() {
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  for (; q--;) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, --r;
    cout << query(l, r) << endl;
  }
}
int main() {
  solve();
  return 0;
}

```