#include <bits/stdc++.h>
using namespace std;
long long mul(long long a, long long b) { return (a * b) % (100000007); }
long long add(long long a, long long b) { return (a + b) % (100000007); }
long long sub(long long a, long long b) {
  return (a - b + llabs(a - b) / (100000007) * (100000007) + (100000007)) %
         (100000007);
}
void upd(long long &a, long long b) {
  a = (a % (100000007) + b % (100000007)) % (100000007);
}
set<int> S;
set<int>::iterator it;
int n, q, a[(100000 + 10)], b[(100000 + 10)];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
pair<int, int> t[(100000 + 10)], ask[(100000 + 10)];
long long ans[(100000 + 10)] = {0};
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n - 1; i++)
    t[i] = make_pair(b[i] = abs(a[i] - a[i + 1]), i);
  sort(t + 1, t + n);
  S.insert(0);
  S.insert(n);
  for (int i = 1; i <= q; i++) cin >> ask[i].first >> ask[i].second;
  for (int i = n - 1; i; i--) {
    int p = t[i].first, m = t[i].second;
    it = S.upper_bound(m);
    int r = (*it);
    it--;
    int l = 1 + (*it);
    S.insert(m);
    for (int j = 1; j <= q; j++) {
      int l2 = max(l, ask[j].first), r2 = min(r, ask[j].second);
      if (l2 <= m && m < r2) ans[j] += (long long)(m - l2 + 1) * (r2 - m) * p;
    }
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long mul(long long a, long long b) { return (a * b) % (100000007); }
long long add(long long a, long long b) { return (a + b) % (100000007); }
long long sub(long long a, long long b) {
  return (a - b + llabs(a - b) / (100000007) * (100000007) + (100000007)) %
         (100000007);
}
void upd(long long &a, long long b) {
  a = (a % (100000007) + b % (100000007)) % (100000007);
}
set<int> S;
set<int>::iterator it;
int n, q, a[(100000 + 10)], b[(100000 + 10)];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
pair<int, int> t[(100000 + 10)], ask[(100000 + 10)];
long long ans[(100000 + 10)] = {0};
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n - 1; i++)
    t[i] = make_pair(b[i] = abs(a[i] - a[i + 1]), i);
  sort(t + 1, t + n);
  S.insert(0);
  S.insert(n);
  for (int i = 1; i <= q; i++) cin >> ask[i].first >> ask[i].second;
  for (int i = n - 1; i; i--) {
    int p = t[i].first, m = t[i].second;
    it = S.upper_bound(m);
    int r = (*it);
    it--;
    int l = 1 + (*it);
    S.insert(m);
    for (int j = 1; j <= q; j++) {
      int l2 = max(l, ask[j].first), r2 = min(r, ask[j].second);
      if (l2 <= m && m < r2) ans[j] += (long long)(m - l2 + 1) * (r2 - m) * p;
    }
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << endl;
  return 0;
}

```