#include <bits/stdc++.h>
using namespace std;
const int SIZE = 100005;
int N, Q;
struct p {
  int v, pos;
} A[SIZE];
int cmp(const p &a, const p &b) { return a.v > b.v; }
int ql[105], qr[105];
set<int> S;
long long res[105];
int lleft[SIZE];
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i].v);
  }
  for (int i = 0; i < N - 1; i++) {
    A[i].v = (A[i].v - A[i + 1].v);
    if (A[i].v < 0) A[i].v = -A[i].v;
    A[i].pos = i + 1;
  }
  sort(A, A + N - 1, cmp);
  for (int i = 1; i <= Q; i++) {
    scanf("%d %d", &ql[i], &qr[i]);
    if (ql[i] > qr[i]) swap(ql[i], qr[i]);
  }
  S.insert(0);
  S.insert(N);
  lleft[N] = 0;
  for (int i = 0; i < N - 1; i++) {
    int l, r;
    int t = A[i].pos;
    r = *S.upper_bound(t);
    l = lleft[r];
    S.insert(t);
    lleft[r] = t;
    lleft[t] = l;
    l++;
    for (int j = 1; j <= Q; j++) {
      int ll = max(l, ql[j]);
      int rr = min(r, qr[j]);
      if (ll <= t && t < rr)
        res[j] += (long long)(t - ll + 1) * (rr - t) * A[i].v;
    }
  }
  for (int i = 1; i <= Q; i++) printf("%I64d\n", res[i]);
  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;
const int SIZE = 100005;
int N, Q;
struct p {
  int v, pos;
} A[SIZE];
int cmp(const p &a, const p &b) { return a.v > b.v; }
int ql[105], qr[105];
set<int> S;
long long res[105];
int lleft[SIZE];
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i].v);
  }
  for (int i = 0; i < N - 1; i++) {
    A[i].v = (A[i].v - A[i + 1].v);
    if (A[i].v < 0) A[i].v = -A[i].v;
    A[i].pos = i + 1;
  }
  sort(A, A + N - 1, cmp);
  for (int i = 1; i <= Q; i++) {
    scanf("%d %d", &ql[i], &qr[i]);
    if (ql[i] > qr[i]) swap(ql[i], qr[i]);
  }
  S.insert(0);
  S.insert(N);
  lleft[N] = 0;
  for (int i = 0; i < N - 1; i++) {
    int l, r;
    int t = A[i].pos;
    r = *S.upper_bound(t);
    l = lleft[r];
    S.insert(t);
    lleft[r] = t;
    lleft[t] = l;
    l++;
    for (int j = 1; j <= Q; j++) {
      int ll = max(l, ql[j]);
      int rr = min(r, qr[j]);
      if (ll <= t && t < rr)
        res[j] += (long long)(t - ll + 1) * (rr - t) * A[i].v;
    }
  }
  for (int i = 1; i <= Q; i++) printf("%I64d\n", res[i]);
  return 0;
}

```