#include <bits/stdc++.h>
using namespace std;
const long long N = 200010;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch - '0'), ch = getchar();
  return s * w;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
long long n, K, a[N], f[N], ql[N], qr[N], sta[N], tp;
signed main() {
  n = read(), K = read();
  for (register long long i = 1; i <= n; i++)
    a[i] = read(), f[i] = abs(a[i] - a[i - 1]);
  for (register long long i = 2; i <= n; i++) {
    while (tp && f[sta[tp]] <= f[i]) qr[sta[tp--]] = i - 1;
    ql[i] = sta[tp];
    sta[++tp] = i;
  }
  while (tp) qr[sta[tp--]] = n;
  for (register long long i = 1; i <= K; i++) {
    long long L, R, res;
    res = 0;
    L = read(), R = read();
    for (register long long j = L + 1; j <= R; j++)
      res += f[j] * (j - max(L, ql[j])) * (min(R, qr[j]) - j + 1);
    printf("%lld\n", res);
  }
  return 0;
}

### Prompt

Generate 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 long long N = 200010;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch - '0'), ch = getchar();
  return s * w;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
long long n, K, a[N], f[N], ql[N], qr[N], sta[N], tp;
signed main() {
  n = read(), K = read();
  for (register long long i = 1; i <= n; i++)
    a[i] = read(), f[i] = abs(a[i] - a[i - 1]);
  for (register long long i = 2; i <= n; i++) {
    while (tp && f[sta[tp]] <= f[i]) qr[sta[tp--]] = i - 1;
    ql[i] = sta[tp];
    sta[++tp] = i;
  }
  while (tp) qr[sta[tp--]] = n;
  for (register long long i = 1; i <= K; i++) {
    long long L, R, res;
    res = 0;
    L = read(), R = read();
    for (register long long j = L + 1; j <= R; j++)
      res += f[j] * (j - max(L, ql[j])) * (min(R, qr[j]) - j + 1);
    printf("%lld\n", res);
  }
  return 0;
}

```