#include <bits/stdc++.h>
using namespace std;
int a[100005], n;
pair<int, int> second[400020];
map<pair<int, int>, long long> all;
void build(int id = 1, int l = 0, int r = n) {
  if (r - l < 2) {
    second[id] = {a[l], l};
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid, r);
  second[id] = max(second[id * 2], second[id * 2 + 1]);
}
pair<int, int> sum(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r || l >= y) return {0, -1};
  if (x <= l && r <= y) return second[id];
  int mid = (l + r) / 2;
  return max(sum(x, y, id * 2, l, mid), sum(x, y, id * 2 + 1, mid, r));
}
int ara[100005];
long long ans;
void do_it(int l, int r) {
  if (l == r) {
    ans += a[l];
    return;
  }
  if (l > r) return;
  pair<int, int> now = sum(l, r + 1);
  long long temp = (now.second - l + 1);
  temp *= (r - now.second + 1);
  temp *= now.first;
  ans += temp;
  do_it(l, now.second - 1);
  do_it(now.second + 1, r);
  return;
}
int main() {
  int N, i, q;
  scanf("%d", &N);
  scanf("%d", &q);
  for (i = 0; i < N; i++) scanf("%d", &ara[i]);
  for (i = 0; i < N - 1; i++) a[i] = abs(ara[i] - ara[i + 1]);
  n = N - 1;
  build();
  for (i = 0; i < q; i++) {
    ans = 0;
    int l, r;
    scanf("%d", &l);
    scanf("%d", &r);
    l--;
    r -= 2;
    if (all.find(make_pair(l, r)) != all.end()) {
      printf("%I64d\n", all[make_pair(l, r)]);
      continue;
    }
    do_it(l, r);
    all[make_pair(l, r)] = ans;
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 a[100005], n;
pair<int, int> second[400020];
map<pair<int, int>, long long> all;
void build(int id = 1, int l = 0, int r = n) {
  if (r - l < 2) {
    second[id] = {a[l], l};
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid, r);
  second[id] = max(second[id * 2], second[id * 2 + 1]);
}
pair<int, int> sum(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r || l >= y) return {0, -1};
  if (x <= l && r <= y) return second[id];
  int mid = (l + r) / 2;
  return max(sum(x, y, id * 2, l, mid), sum(x, y, id * 2 + 1, mid, r));
}
int ara[100005];
long long ans;
void do_it(int l, int r) {
  if (l == r) {
    ans += a[l];
    return;
  }
  if (l > r) return;
  pair<int, int> now = sum(l, r + 1);
  long long temp = (now.second - l + 1);
  temp *= (r - now.second + 1);
  temp *= now.first;
  ans += temp;
  do_it(l, now.second - 1);
  do_it(now.second + 1, r);
  return;
}
int main() {
  int N, i, q;
  scanf("%d", &N);
  scanf("%d", &q);
  for (i = 0; i < N; i++) scanf("%d", &ara[i]);
  for (i = 0; i < N - 1; i++) a[i] = abs(ara[i] - ara[i + 1]);
  n = N - 1;
  build();
  for (i = 0; i < q; i++) {
    ans = 0;
    int l, r;
    scanf("%d", &l);
    scanf("%d", &r);
    l--;
    r -= 2;
    if (all.find(make_pair(l, r)) != all.end()) {
      printf("%I64d\n", all[make_pair(l, r)]);
      continue;
    }
    do_it(l, r);
    all[make_pair(l, r)] = ans;
    printf("%I64d\n", ans);
  }
  return 0;
}

```