#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int SIZE = 3e5 + 5;
const long long INF = 1000000000000000000LL;
int rd[100009];
long long d[100009];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < (n); ++i) scanf("%d", &(rd[i]));
  for (int i = 0; i < (n - 1); ++i) {
    d[i] = abs(rd[i + 1] - rd[i]);
  }
  for (int it = 0; it < (q); ++it) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    if (a == b) {
      printf("0\n");
    }
    vector<pair<long long, long long> > v;
    long long ans = 0;
    long long cv = 0;
    v.push_back(make_pair(1 << 30, a - 1));
    for (int i = (a); i < (b); ++i) {
      while (v.back().first <= d[i]) {
        if (v.size() > 1)
          cv -= (long long)v.back().first *
                (long long)(v[v.size() - 1].second - v[v.size() - 2].second);
        v.pop_back();
      }
      v.push_back(make_pair(d[i], i));
      if (v.size() > 1)
        cv += (long long)v.back().first *
              (long long)(v[v.size() - 1].second - v[v.size() - 2].second);
      ans += cv;
    }
    printf("%I64d\n", ans);
  }
}

### Prompt

Create a solution in cpp for 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 MOD = 1e9 + 7;
const int SIZE = 3e5 + 5;
const long long INF = 1000000000000000000LL;
int rd[100009];
long long d[100009];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < (n); ++i) scanf("%d", &(rd[i]));
  for (int i = 0; i < (n - 1); ++i) {
    d[i] = abs(rd[i + 1] - rd[i]);
  }
  for (int it = 0; it < (q); ++it) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    if (a == b) {
      printf("0\n");
    }
    vector<pair<long long, long long> > v;
    long long ans = 0;
    long long cv = 0;
    v.push_back(make_pair(1 << 30, a - 1));
    for (int i = (a); i < (b); ++i) {
      while (v.back().first <= d[i]) {
        if (v.size() > 1)
          cv -= (long long)v.back().first *
                (long long)(v[v.size() - 1].second - v[v.size() - 2].second);
        v.pop_back();
      }
      v.push_back(make_pair(d[i], i));
      if (v.size() > 1)
        cv += (long long)v.back().first *
              (long long)(v[v.size() - 1].second - v[v.size() - 2].second);
      ans += cv;
    }
    printf("%I64d\n", ans);
  }
}

```