#include <bits/stdc++.h>
const int MAXN = 1e5 + 5;
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s < '0' || '9' < s) {
    f = 1;
    if (s == '-') f = -1;
    s = getchar();
  }
  while ('0' <= s && s <= '9') {
    x = (x << 3) + (x << 1) + s - '0', s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) {
    putchar('-'), x = -x;
  }
  if (9 < x) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename _T>
_T ABS(const _T a) {
  return a < 0 ? -a : a;
}
int sta[MAXN], top;
int a[MAXN], K[MAXN];
int N, M;
int main() {
  read(N), read(M);
  for (int i = 1; i <= N; i++) read(a[i]);
  for (int i = 2; i <= N; i++) K[i] = ABS(a[i] - a[i - 1]);
  long long res, tmp;
  int l, r;
  while (M--) {
    read(l), read(r);
    res = 0, tmp = 0;
    sta[top = 0] = l;
    for (int i = l + 1; i <= r; i++) {
      while (top && K[sta[top]] <= K[i])
        tmp -= 1ll * (sta[top] - sta[top - 1]) * K[sta[top]], top--;
      tmp += 1ll * (i - sta[top]) * K[i], sta[++top] = i;
      res += tmp;
    }
    write(res), putchar('\n');
  }
  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>
const int MAXN = 1e5 + 5;
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s < '0' || '9' < s) {
    f = 1;
    if (s == '-') f = -1;
    s = getchar();
  }
  while ('0' <= s && s <= '9') {
    x = (x << 3) + (x << 1) + s - '0', s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) {
    putchar('-'), x = -x;
  }
  if (9 < x) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename _T>
_T ABS(const _T a) {
  return a < 0 ? -a : a;
}
int sta[MAXN], top;
int a[MAXN], K[MAXN];
int N, M;
int main() {
  read(N), read(M);
  for (int i = 1; i <= N; i++) read(a[i]);
  for (int i = 2; i <= N; i++) K[i] = ABS(a[i] - a[i - 1]);
  long long res, tmp;
  int l, r;
  while (M--) {
    read(l), read(r);
    res = 0, tmp = 0;
    sta[top = 0] = l;
    for (int i = l + 1; i <= r; i++) {
      while (top && K[sta[top]] <= K[i])
        tmp -= 1ll * (sta[top] - sta[top - 1]) * K[sta[top]], top--;
      tmp += 1ll * (i - sta[top]) * K[i], sta[++top] = i;
      res += tmp;
    }
    write(res), putchar('\n');
  }
  return 0;
}

```