#include <bits/stdc++.h>
template <typename Tp>
Tp Max(const Tp &a, const Tp &b) {
  return a > b ? a : b;
}
template <typename Tp>
Tp Min(const Tp &a, const Tp &b) {
  return a < b ? a : b;
}
template <typename Tp>
Tp Abs(const Tp &a) {
  return a > 0 ? a : -a;
}
template <typename Tp>
void Read(Tp &x) {
  Tp in = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    in = in * 10 + ch - '0';
    ch = getchar();
  }
  x = in * f;
}
const int SN = 100000 + 10;
long long a[SN], pos[SN], sta[SN], top, tot, n;
long long left[SN], right[SN], d[SN];
int L, R;
void Answer() {
  top = 0;
  right[pos[1]] = pos[tot];
  sta[++top] = pos[1];
  for (int i = 2; i <= tot; i++) {
    while (top && d[pos[i]] > d[sta[top]]) right[sta[top--]] = pos[i] - 1;
    sta[++top] = pos[i], right[pos[i]] = pos[tot];
  }
  top = 0;
  left[pos[tot]] = pos[1];
  sta[++top] = pos[tot];
  for (int i = tot - 1; i >= 1; i--) {
    while (top && d[pos[i]] >= d[sta[top]]) left[sta[top--]] = pos[i] + 1;
    sta[++top] = pos[i], left[pos[i]] = pos[1];
  }
  long long ans = 0;
  for (int i = L; i < R; i++) {
    ans = (ans + d[i] * (i - left[i] + 1) * (right[i] - i + 1));
  }
  printf("%I64d\n", ans);
}
int main() {
  int q;
  Read(n), Read(q);
  for (int i = 1; i <= n; i++) Read(a[i]);
  for (int i = 1; i < n; i++) d[i] = Abs(a[i] - a[i + 1]);
  while (q--) {
    Read(L), Read(R);
    tot = 0;
    for (int i = L; i < R; i++) pos[++tot] = i;
    pos[tot + 1] = 0;
    Answer();
  }
  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>
template <typename Tp>
Tp Max(const Tp &a, const Tp &b) {
  return a > b ? a : b;
}
template <typename Tp>
Tp Min(const Tp &a, const Tp &b) {
  return a < b ? a : b;
}
template <typename Tp>
Tp Abs(const Tp &a) {
  return a > 0 ? a : -a;
}
template <typename Tp>
void Read(Tp &x) {
  Tp in = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    in = in * 10 + ch - '0';
    ch = getchar();
  }
  x = in * f;
}
const int SN = 100000 + 10;
long long a[SN], pos[SN], sta[SN], top, tot, n;
long long left[SN], right[SN], d[SN];
int L, R;
void Answer() {
  top = 0;
  right[pos[1]] = pos[tot];
  sta[++top] = pos[1];
  for (int i = 2; i <= tot; i++) {
    while (top && d[pos[i]] > d[sta[top]]) right[sta[top--]] = pos[i] - 1;
    sta[++top] = pos[i], right[pos[i]] = pos[tot];
  }
  top = 0;
  left[pos[tot]] = pos[1];
  sta[++top] = pos[tot];
  for (int i = tot - 1; i >= 1; i--) {
    while (top && d[pos[i]] >= d[sta[top]]) left[sta[top--]] = pos[i] + 1;
    sta[++top] = pos[i], left[pos[i]] = pos[1];
  }
  long long ans = 0;
  for (int i = L; i < R; i++) {
    ans = (ans + d[i] * (i - left[i] + 1) * (right[i] - i + 1));
  }
  printf("%I64d\n", ans);
}
int main() {
  int q;
  Read(n), Read(q);
  for (int i = 1; i <= n; i++) Read(a[i]);
  for (int i = 1; i < n; i++) d[i] = Abs(a[i] - a[i + 1]);
  while (q--) {
    Read(L), Read(R);
    tot = 0;
    for (int i = L; i < R; i++) pos[++tot] = i;
    pos[tot + 1] = 0;
    Answer();
  }
  return 0;
}

```