#include <bits/stdc++.h>
using namespace std;
const int mxint = numeric_limits<int>::max();
const int mnint = numeric_limits<int>::min();
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 100010;
int n, q;
int a[maxn];
int b[maxn];
int l[maxn], r[maxn];
struct Node {
  int b, p;
} s[maxn];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (i > 1) b[i - 1] = abs(a[i] - a[i - 1]);
  }
  b[0] = b[n] = inf;
  int p = 0;
  s[p++] = Node{inf, 0};
  for (int i = 1; i < n; i++) {
    for (int j = p - 1; j >= 0; j--) {
      if (b[i] <= s[j].b) {
        l[i] = s[j].p;
        p = j + 1;
        s[p++] = Node{b[i], i};
        break;
      }
    }
  }
  p = 0;
  s[p++] = Node{inf, n};
  for (int i = n - 1; i >= 1; i--) {
    for (int j = p - 1; j >= 0; j--) {
      if (b[i] < s[j].b) {
        r[i] = s[j].p;
        p = j + 1;
        s[p++] = Node{b[i], i};
        break;
      }
    }
  }
  while (q--) {
    int ll, rr;
    scanf("%d%d", &ll, &rr);
    long long ans = 0;
    for (int i = ll; i < rr; i++) {
      int tl = max(ll - 1, l[i]);
      int tr = min(rr, r[i]);
      ans += (long long)(i - tl) * (tr - i) * b[i];
    }
    printf("%I64d\n", ans);
  }
}

### Prompt

Create a solution in CPP for 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 mxint = numeric_limits<int>::max();
const int mnint = numeric_limits<int>::min();
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 100010;
int n, q;
int a[maxn];
int b[maxn];
int l[maxn], r[maxn];
struct Node {
  int b, p;
} s[maxn];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (i > 1) b[i - 1] = abs(a[i] - a[i - 1]);
  }
  b[0] = b[n] = inf;
  int p = 0;
  s[p++] = Node{inf, 0};
  for (int i = 1; i < n; i++) {
    for (int j = p - 1; j >= 0; j--) {
      if (b[i] <= s[j].b) {
        l[i] = s[j].p;
        p = j + 1;
        s[p++] = Node{b[i], i};
        break;
      }
    }
  }
  p = 0;
  s[p++] = Node{inf, n};
  for (int i = n - 1; i >= 1; i--) {
    for (int j = p - 1; j >= 0; j--) {
      if (b[i] < s[j].b) {
        r[i] = s[j].p;
        p = j + 1;
        s[p++] = Node{b[i], i};
        break;
      }
    }
  }
  while (q--) {
    int ll, rr;
    scanf("%d%d", &ll, &rr);
    long long ans = 0;
    for (int i = ll; i < rr; i++) {
      int tl = max(ll - 1, l[i]);
      int tr = min(rr, r[i]);
      ans += (long long)(i - tl) * (tr - i) * b[i];
    }
    printf("%I64d\n", ans);
  }
}

```