#include <bits/stdc++.h>
using namespace std;
int n, m;
int v[100010], id[100010];
int ab(int x) {
  if (x < 0) return -x;
  return x;
}
inline int cmp(int a, int b) { return v[a] < v[b]; }
int t[100010], f[100010], mn[100010], mx[100010];
int tata(int nod) {
  if (t[nod] == nod) return nod;
  int aux = tata(t[nod]);
  t[nod] = aux;
  return aux;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);
  for (int i = 1; i < n; ++i) v[i] = ab(v[i] - v[i + 1]);
  for (int i = 1; i < n; ++i) id[i] = i;
  sort(id + 1, id + n, cmp);
  while (m--) {
    int a, b;
    scanf("%d%d", &a, &b);
    b--;
    memset(f, 0, sizeof(f));
    long long sol = 0;
    for (int i = 1; i < n; ++i) {
      if (id[i] < a || id[i] > b) continue;
      int nod = id[i];
      f[nod] = 1;
      t[nod] = nod;
      mx[nod] = mn[nod] = nod;
      int t1, t2;
      if (f[nod - 1]) {
        t1 = tata(nod - 1);
        t2 = tata(nod);
        t[t1] = t2;
        mn[t2] = min(mn[t1], mn[t2]);
        mx[t2] = max(mx[t1], mx[t2]);
      }
      if (f[nod + 1]) {
        t1 = tata(nod + 1);
        t2 = tata(nod);
        t[t1] = t2;
        mn[t2] = min(mn[t1], mn[t2]);
        mx[t2] = max(mx[t1], mx[t2]);
      }
      sol = sol + (1LL * nod - mn[nod] + 1) * (mx[nod] - nod + 1) * v[id[i]];
    }
    printf("%I64d\n", sol);
  }
  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;
int n, m;
int v[100010], id[100010];
int ab(int x) {
  if (x < 0) return -x;
  return x;
}
inline int cmp(int a, int b) { return v[a] < v[b]; }
int t[100010], f[100010], mn[100010], mx[100010];
int tata(int nod) {
  if (t[nod] == nod) return nod;
  int aux = tata(t[nod]);
  t[nod] = aux;
  return aux;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);
  for (int i = 1; i < n; ++i) v[i] = ab(v[i] - v[i + 1]);
  for (int i = 1; i < n; ++i) id[i] = i;
  sort(id + 1, id + n, cmp);
  while (m--) {
    int a, b;
    scanf("%d%d", &a, &b);
    b--;
    memset(f, 0, sizeof(f));
    long long sol = 0;
    for (int i = 1; i < n; ++i) {
      if (id[i] < a || id[i] > b) continue;
      int nod = id[i];
      f[nod] = 1;
      t[nod] = nod;
      mx[nod] = mn[nod] = nod;
      int t1, t2;
      if (f[nod - 1]) {
        t1 = tata(nod - 1);
        t2 = tata(nod);
        t[t1] = t2;
        mn[t2] = min(mn[t1], mn[t2]);
        mx[t2] = max(mx[t1], mx[t2]);
      }
      if (f[nod + 1]) {
        t1 = tata(nod + 1);
        t2 = tata(nod);
        t[t1] = t2;
        mn[t2] = min(mn[t1], mn[t2]);
        mx[t2] = max(mx[t1], mx[t2]);
      }
      sol = sol + (1LL * nod - mn[nod] + 1) * (mx[nod] - nod + 1) * v[id[i]];
    }
    printf("%I64d\n", sol);
  }
  return 0;
}

```