#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T& n) {
  char ch;
  int sign = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = -1;
  n = ch - '0';
  while (isdigit(ch = getchar())) n = n * 10 + ch - '0';
  n *= sign;
}
const int INF = 1e9 + 7;
const int N = 122222;
int n, m, l, r;
int stk[N], tp;
long long a[N], le[N], ri[N];
int main() {
  read(n), read(m);
  for (int i = (1); i <= (n); ++i) read(a[i]);
  for (int i = (1); i <= (n - 1); ++i) a[i] = abs(a[i] - a[i + 1]);
  a[0] = a[n + 1] = INF;
  while (m--) {
    read(l), read(r);
    stk[tp = 1] = 0;
    for (int i = (l); i <= (r - 1); ++i) {
      for (; tp > 1 && a[stk[tp]] < a[i]; tp--)
        ;
      le[i] = max(l - 1, stk[tp]), stk[++tp] = i;
    }
    stk[tp = 1] = n + 1;
    for (int i = (r - 1); i >= (l); --i) {
      for (; tp > 1 && a[stk[tp]] <= a[i]; tp--)
        ;
      ri[i] = min(r, stk[tp]), stk[++tp] = i;
    }
    long long ans = 0;
    for (long long i = l; i < r; ++i) {
      long long first = (ri[i] - i) * (i - le[i]);
      ans += first * a[i];
    }
    cout << ans << endl;
  }
  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>
using namespace std;
template <typename T>
void read(T& n) {
  char ch;
  int sign = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = -1;
  n = ch - '0';
  while (isdigit(ch = getchar())) n = n * 10 + ch - '0';
  n *= sign;
}
const int INF = 1e9 + 7;
const int N = 122222;
int n, m, l, r;
int stk[N], tp;
long long a[N], le[N], ri[N];
int main() {
  read(n), read(m);
  for (int i = (1); i <= (n); ++i) read(a[i]);
  for (int i = (1); i <= (n - 1); ++i) a[i] = abs(a[i] - a[i + 1]);
  a[0] = a[n + 1] = INF;
  while (m--) {
    read(l), read(r);
    stk[tp = 1] = 0;
    for (int i = (l); i <= (r - 1); ++i) {
      for (; tp > 1 && a[stk[tp]] < a[i]; tp--)
        ;
      le[i] = max(l - 1, stk[tp]), stk[++tp] = i;
    }
    stk[tp = 1] = n + 1;
    for (int i = (r - 1); i >= (l); --i) {
      for (; tp > 1 && a[stk[tp]] <= a[i]; tp--)
        ;
      ri[i] = min(r, stk[tp]), stk[++tp] = i;
    }
    long long ans = 0;
    for (long long i = l; i < r; ++i) {
      long long first = (ri[i] - i) * (i - le[i]);
      ans += first * a[i];
    }
    cout << ans << endl;
  }
  return 0;
}

```