#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int inf = 1061109567;
const int dir[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
const int maxn = 1e6 + 5;
int a[maxn], d[maxn], l[maxn], r[maxn], q[maxn];
int main() {
  int n, qe, L, R;
  cin >> n >> qe;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i < n; i++) {
    d[i] = abs(a[i] - a[i - 1]);
  }
  while (qe--) {
    scanf("%d%d", &L, &R);
    int pos = 0;
    for (int i = L; i < R; i++) {
      while (pos && d[q[pos - 1]] <= d[i]) {
        r[q[pos - 1]] = i - 1;
        pos--;
      }
      q[pos++] = i;
    }
    while (pos) {
      r[q[pos - 1]] = R - 1;
      pos--;
    }
    for (int i = R - 1; i >= L; i--) {
      while (pos && d[q[pos - 1]] < d[i]) {
        l[q[pos - 1]] = i + 1;
        pos--;
      }
      q[pos++] = i;
    }
    while (pos) {
      l[q[pos - 1]] = L;
      pos--;
    }
    long long ans = 0;
    for (int i = L; i < R; i++) {
      ans += 1ll * (i - l[i] + 1) * (1 + r[i] - i) * d[i];
    }
    cout << ans << endl;
  }
}

### 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;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int inf = 1061109567;
const int dir[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
const int maxn = 1e6 + 5;
int a[maxn], d[maxn], l[maxn], r[maxn], q[maxn];
int main() {
  int n, qe, L, R;
  cin >> n >> qe;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i < n; i++) {
    d[i] = abs(a[i] - a[i - 1]);
  }
  while (qe--) {
    scanf("%d%d", &L, &R);
    int pos = 0;
    for (int i = L; i < R; i++) {
      while (pos && d[q[pos - 1]] <= d[i]) {
        r[q[pos - 1]] = i - 1;
        pos--;
      }
      q[pos++] = i;
    }
    while (pos) {
      r[q[pos - 1]] = R - 1;
      pos--;
    }
    for (int i = R - 1; i >= L; i--) {
      while (pos && d[q[pos - 1]] < d[i]) {
        l[q[pos - 1]] = i + 1;
        pos--;
      }
      q[pos++] = i;
    }
    while (pos) {
      l[q[pos - 1]] = L;
      pos--;
    }
    long long ans = 0;
    for (int i = L; i < R; i++) {
      ans += 1ll * (i - l[i] + 1) * (1 + r[i] - i) * d[i];
    }
    cout << ans << endl;
  }
}

```