#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 15;
const int Q = 1e9 + 7;
int a[M], L[M], R[M], st[M];
int h[M], hh[M];
long long ans(int n) {
  int r;
  a[0] = Q, a[n + 1] = Q;
  r = 1;
  st[0] = 0;
  for (int i = 1; i <= n; i++) {
    while (a[st[r - 1]] <= a[i]) {
      r--;
    }
    L[i] = st[r - 1];
    st[r++] = i;
  }
  st[0] = n + 1;
  r = 1;
  for (int i = n; i >= 1; i--) {
    while (a[st[r - 1]] < a[i]) {
      r--;
    }
    R[i] = st[r - 1];
    st[r++] = i;
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += 1LL * a[i] * (i - L[i]) * (R[i] - i);
  }
  return res;
}
int main() {
  srand(time(NULL));
  int n;
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    scanf("%d", &h[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    hh[i] = abs(h[i] - h[i + 1]);
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    for (int j = 0; j < r - l; j++) {
      a[j + 1] = hh[l + j - 1];
    }
    cout << ans(r - l) << "\n";
  }
  cerr << clock() / (CLOCKS_PER_SEC + 0.) << endl;
  return 0;
}

### Prompt

Please formulate 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 int M = 1e5 + 15;
const int Q = 1e9 + 7;
int a[M], L[M], R[M], st[M];
int h[M], hh[M];
long long ans(int n) {
  int r;
  a[0] = Q, a[n + 1] = Q;
  r = 1;
  st[0] = 0;
  for (int i = 1; i <= n; i++) {
    while (a[st[r - 1]] <= a[i]) {
      r--;
    }
    L[i] = st[r - 1];
    st[r++] = i;
  }
  st[0] = n + 1;
  r = 1;
  for (int i = n; i >= 1; i--) {
    while (a[st[r - 1]] < a[i]) {
      r--;
    }
    R[i] = st[r - 1];
    st[r++] = i;
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += 1LL * a[i] * (i - L[i]) * (R[i] - i);
  }
  return res;
}
int main() {
  srand(time(NULL));
  int n;
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    scanf("%d", &h[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    hh[i] = abs(h[i] - h[i + 1]);
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    for (int j = 0; j < r - l; j++) {
      a[j + 1] = hh[l + j - 1];
    }
    cout << ans(r - l) << "\n";
  }
  cerr << clock() / (CLOCKS_PER_SEC + 0.) << endl;
  return 0;
}

```