#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 9, maxn = 100000;
int L[maxn + 99], R[maxn + 99], h[maxn + 99], Stack[maxn + 99], pos[maxn + 99],
    use[maxn + 9];
int n, q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &h[i]);
  int top = 0;
  Stack[0] = inf;
  pos[0] = 0;
  for (int i = 1; i < n; ++i) {
    int value = abs(h[i] - h[i + 1]);
    use[i] = value;
    while (top > 0 && Stack[top] < value) --top;
    L[i] = i - pos[top];
    pos[++top] = i;
    Stack[top] = value;
  }
  top = 0;
  Stack[0] = inf;
  pos[0] = n;
  for (int i = n - 1; i >= 1; --i) {
    int value = abs(h[i] - h[i + 1]);
    while (top > 0 && Stack[top] <= value) --top;
    R[i] = pos[top] - i;
    pos[++top] = i;
    Stack[top] = value;
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    --r;
    long long ans = 0;
    for (int i = l; i <= r; ++i) {
      int lq = max(l - 1, i - L[i]) + 1, rq = min(r + 1, i + R[i]) - 1;
      ans += use[i] * (rq - i + 1LL) * (i - lq + 1LL);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### 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 int inf = 1e9 + 9, maxn = 100000;
int L[maxn + 99], R[maxn + 99], h[maxn + 99], Stack[maxn + 99], pos[maxn + 99],
    use[maxn + 9];
int n, q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &h[i]);
  int top = 0;
  Stack[0] = inf;
  pos[0] = 0;
  for (int i = 1; i < n; ++i) {
    int value = abs(h[i] - h[i + 1]);
    use[i] = value;
    while (top > 0 && Stack[top] < value) --top;
    L[i] = i - pos[top];
    pos[++top] = i;
    Stack[top] = value;
  }
  top = 0;
  Stack[0] = inf;
  pos[0] = n;
  for (int i = n - 1; i >= 1; --i) {
    int value = abs(h[i] - h[i + 1]);
    while (top > 0 && Stack[top] <= value) --top;
    R[i] = pos[top] - i;
    pos[++top] = i;
    Stack[top] = value;
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    --r;
    long long ans = 0;
    for (int i = l; i <= r; ++i) {
      int lq = max(l - 1, i - L[i]) + 1, rq = min(r + 1, i + R[i]) - 1;
      ans += use[i] * (rq - i + 1LL) * (i - lq + 1LL);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```