#include <bits/stdc++.h>
using namespace std;
const int lim = 1e5 + 11;
int f[lim];
int s[lim];
int qq;
int x, y;
int ans;
int m, n;
int l[lim], r[lim];
int pos, old;
void add(int o, int l, int r) {
  if (l == r) {
    f[o] = old;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    add(o << 1, l, mid);
  else
    add(o << 1 | 1, mid + 1, r);
  f[o] = max(f[o << 1], f[o << 1 | 1]);
}
void work() {
  unsigned long long ans = 0;
  for (int i = x + 1; i <= y; i++) {
    ans += ((unsigned long long)(i - l[i]) * (r[i] - i) * f[i]);
  }
  cout << ans << endl;
}
void fl(int now, int k) {
  if (k <= x || f[k] >= f[now]) {
    l[now] = k;
    return;
  }
  fl(now, l[k]);
}
void fr(int now, int k) {
  if (k > y || f[k] > f[now]) {
    r[now] = k;
    return;
  }
  fr(now, r[k]);
}
int main() {
  scanf("%d%d", &m, &qq);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &s[i]);
  }
  for (int i = 1; i <= qq; i++) {
    scanf("%d%d", &x, &y);
    n = 0;
    for (int i = x + 1; i <= y; i++) f[i] = abs(s[i] - s[i - 1]);
    l[x + 1] = x;
    for (int i = x + 2; i <= y; i++) {
      if (f[i] <= f[i - 1])
        l[i] = i - 1;
      else
        fl(i, l[i - 1]);
    }
    r[y] = y + 1;
    for (int i = y - 1; i > x; i--) {
      if (f[i] < f[i + 1])
        r[i] = i + 1;
      else
        fr(i, r[i + 1]);
    }
    work();
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 lim = 1e5 + 11;
int f[lim];
int s[lim];
int qq;
int x, y;
int ans;
int m, n;
int l[lim], r[lim];
int pos, old;
void add(int o, int l, int r) {
  if (l == r) {
    f[o] = old;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    add(o << 1, l, mid);
  else
    add(o << 1 | 1, mid + 1, r);
  f[o] = max(f[o << 1], f[o << 1 | 1]);
}
void work() {
  unsigned long long ans = 0;
  for (int i = x + 1; i <= y; i++) {
    ans += ((unsigned long long)(i - l[i]) * (r[i] - i) * f[i]);
  }
  cout << ans << endl;
}
void fl(int now, int k) {
  if (k <= x || f[k] >= f[now]) {
    l[now] = k;
    return;
  }
  fl(now, l[k]);
}
void fr(int now, int k) {
  if (k > y || f[k] > f[now]) {
    r[now] = k;
    return;
  }
  fr(now, r[k]);
}
int main() {
  scanf("%d%d", &m, &qq);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &s[i]);
  }
  for (int i = 1; i <= qq; i++) {
    scanf("%d%d", &x, &y);
    n = 0;
    for (int i = x + 1; i <= y; i++) f[i] = abs(s[i] - s[i - 1]);
    l[x + 1] = x;
    for (int i = x + 2; i <= y; i++) {
      if (f[i] <= f[i - 1])
        l[i] = i - 1;
      else
        fl(i, l[i - 1]);
    }
    r[y] = y + 1;
    for (int i = y - 1; i > x; i--) {
      if (f[i] < f[i + 1])
        r[i] = i + 1;
      else
        fr(i, r[i + 1]);
    }
    work();
  }
  return 0;
}

```