#include <bits/stdc++.h>
using namespace std;
const int maxm = 1e6 + 10;
const int maxn = 2e5 + 10;
int n, qq;
int a[maxn], l[maxn], r[maxn], q[maxn], pos[maxn];
int v[maxn];
int main() {
  while (~scanf("%d%d", &n, &qq)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i < n; i++) v[i] = abs(a[i + 1] - a[i]);
    q[0] = 1e9;
    pos[0] = 0;
    int pp = 0;
    for (int i = 1; i < n; i++) {
      while (q[pp] <= v[i]) pp--;
      l[i] = pos[pp];
      q[++pp] = v[i];
      pos[pp] = i;
    }
    q[0] = 1e9;
    pos[0] = n;
    pp = 0;
    for (int i = n - 1; i > 0; i--) {
      while (q[pp] < v[i]) pp--;
      r[i] = pos[pp];
      q[++pp] = v[i];
      pos[pp] = i;
    }
    while (qq--) {
      int ll, rr;
      scanf("%d%d", &ll, &rr);
      long long ans = 0;
      for (int i = ll; i < rr; i++) {
        long long tpv = 1;
        tpv = tpv * (i - max(l[i] + 1, ll) + 1) * (min(r[i], rr) - i) * v[i];
        ans += tpv;
      }
      printf("%I64d\n", ans);
    }
  }
}

### 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 int maxm = 1e6 + 10;
const int maxn = 2e5 + 10;
int n, qq;
int a[maxn], l[maxn], r[maxn], q[maxn], pos[maxn];
int v[maxn];
int main() {
  while (~scanf("%d%d", &n, &qq)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i < n; i++) v[i] = abs(a[i + 1] - a[i]);
    q[0] = 1e9;
    pos[0] = 0;
    int pp = 0;
    for (int i = 1; i < n; i++) {
      while (q[pp] <= v[i]) pp--;
      l[i] = pos[pp];
      q[++pp] = v[i];
      pos[pp] = i;
    }
    q[0] = 1e9;
    pos[0] = n;
    pp = 0;
    for (int i = n - 1; i > 0; i--) {
      while (q[pp] < v[i]) pp--;
      r[i] = pos[pp];
      q[++pp] = v[i];
      pos[pp] = i;
    }
    while (qq--) {
      int ll, rr;
      scanf("%d%d", &ll, &rr);
      long long ans = 0;
      for (int i = ll; i < rr; i++) {
        long long tpv = 1;
        tpv = tpv * (i - max(l[i] + 1, ll) + 1) * (min(r[i], rr) - i) * v[i];
        ans += tpv;
      }
      printf("%I64d\n", ans);
    }
  }
}

```