#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
int n, nq;
int x[MAXN];
pair<int, int> d[MAXN - 1];
int uf[MAXN];
int find(int x) {
  if (uf[x] < 0) return x;
  return uf[x] = find(uf[x]);
}
void unite(int a, int b) {
  if (-uf[a] < -uf[b]) swap(a, b);
  uf[a] += uf[b];
  uf[b] = a;
}
void run() {
  scanf("%d%d", &n, &nq);
  for (int i = (0); i < (n); ++i) scanf("%d", &x[i]);
  for (int i = (0); i < (n - 1); ++i) d[i] = make_pair(abs(x[i + 1] - x[i]), i);
  sort(d, d + n - 1);
  for (int qi = (0); qi < (nq); ++qi) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, --r;
    long long ret = 0;
    for (int i = (0); i < (n); ++i) uf[i] = -1;
    for (int i = (0); i < (n - 1); ++i) {
      int a = d[i].second, b = d[i].second + 1, val = d[i].first;
      if (a < l || b > r) continue;
      a = find(a), b = find(b);
      ret += (long long)uf[a] * uf[b] * val;
      unite(a, b);
    }
    printf("%I64d\n", ret);
  }
}
int main() {
  run();
  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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
int n, nq;
int x[MAXN];
pair<int, int> d[MAXN - 1];
int uf[MAXN];
int find(int x) {
  if (uf[x] < 0) return x;
  return uf[x] = find(uf[x]);
}
void unite(int a, int b) {
  if (-uf[a] < -uf[b]) swap(a, b);
  uf[a] += uf[b];
  uf[b] = a;
}
void run() {
  scanf("%d%d", &n, &nq);
  for (int i = (0); i < (n); ++i) scanf("%d", &x[i]);
  for (int i = (0); i < (n - 1); ++i) d[i] = make_pair(abs(x[i + 1] - x[i]), i);
  sort(d, d + n - 1);
  for (int qi = (0); qi < (nq); ++qi) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, --r;
    long long ret = 0;
    for (int i = (0); i < (n); ++i) uf[i] = -1;
    for (int i = (0); i < (n - 1); ++i) {
      int a = d[i].second, b = d[i].second + 1, val = d[i].first;
      if (a < l || b > r) continue;
      a = find(a), b = find(b);
      ret += (long long)uf[a] * uf[b] * val;
      unite(a, b);
    }
    printf("%I64d\n", ret);
  }
}
int main() {
  run();
  return 0;
}

```