#include <bits/stdc++.h>
using namespace std;
const long long int M = 5e5 + 5, LG = 17, inf = 1e18;
int n, q;
pair<int, int> rmq[M][LG];
long long int lg[M];
pair<int, int> get(long long int l, long long int r) {
  long long int x = lg[r - l];
  return max(rmq[l][x], rmq[r - (1 << x)][x]);
}
long long int calc(long long int l, long long int r) {
  if (l + 1 == r) return rmq[l][0].first;
  if (l == r) return 0;
  pair<int, int> x = get(l, r);
  return x.first * (x.second - l + 1) * (r - x.second) + calc(l, x.second) +
         calc(x.second + 1, r);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 2; i < n; i++) {
    lg[i] = lg[i / 2] + 1;
  }
  n--;
  int pv;
  scanf("%d", &pv);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    rmq[i][0] = {abs(pv - x), i};
    pv = x;
  }
  for (int i = 1; i < LG; i++) {
    for (int j = 0; j < n; j++) {
      rmq[j][i] = max(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
    }
  }
  for (int i = 0; i < q; i++) {
    long long int l, r;
    cin >> l >> r;
    l--;
    r--;
    printf("%lld\n", calc(l, r));
  }
}

### Prompt

Your challenge is to write 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 int M = 5e5 + 5, LG = 17, inf = 1e18;
int n, q;
pair<int, int> rmq[M][LG];
long long int lg[M];
pair<int, int> get(long long int l, long long int r) {
  long long int x = lg[r - l];
  return max(rmq[l][x], rmq[r - (1 << x)][x]);
}
long long int calc(long long int l, long long int r) {
  if (l + 1 == r) return rmq[l][0].first;
  if (l == r) return 0;
  pair<int, int> x = get(l, r);
  return x.first * (x.second - l + 1) * (r - x.second) + calc(l, x.second) +
         calc(x.second + 1, r);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 2; i < n; i++) {
    lg[i] = lg[i / 2] + 1;
  }
  n--;
  int pv;
  scanf("%d", &pv);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    rmq[i][0] = {abs(pv - x), i};
    pv = x;
  }
  for (int i = 1; i < LG; i++) {
    for (int j = 0; j < n; j++) {
      rmq[j][i] = max(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
    }
  }
  for (int i = 0; i < q; i++) {
    long long int l, r;
    cin >> l >> r;
    l--;
    r--;
    printf("%lld\n", calc(l, r));
  }
}

```