#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = (long long int)1e9 + 1;
const int mod = 998244353, sz = 5, MAX = (long long int)1e6 + 10;
string fir = "Monocarp", sec = "Bicarp";
void precom() {}
void result(long long int n, vector<long long int> a) {
  vector<long long int> l(n), r(n);
  for (long long int i = 0; i < n; i++) l[i] = r[i] = i;
  for (long long int i = 1; i < n; i++) {
    long long int st = i;
    while (st && a[i] >= a[st - 1]) {
      st = l[st - 1];
      l[i] = st;
    }
  }
  for (long long int i = n - 2; i >= 0; i--) {
    long long int st = i;
    while (st < n - 1 && a[i] > a[st + 1]) {
      st = r[st + 1];
      r[i] = st;
    }
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans += a[i] * (i - l[i] + 1) * (r[i] - i + 1);
  }
  cout << ans << "\n";
  return;
}
void work() {
  long long int n, q;
  cin >> n >> q;
  vector<long long int> a(N, 0ll);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  vector<long long int> b(n - 1);
  for (long long int i = 0; i + 1 < n; i++) b[i] = abs(a[i] - a[i + 1]);
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    x--, y--;
    result(y - x, vector<long long int>(b.begin() + x, b.begin() + y));
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    work();
  }
  return 0;
}

### Prompt

Your challenge is to write 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 N = 1e5 + 10, M = (long long int)1e9 + 1;
const int mod = 998244353, sz = 5, MAX = (long long int)1e6 + 10;
string fir = "Monocarp", sec = "Bicarp";
void precom() {}
void result(long long int n, vector<long long int> a) {
  vector<long long int> l(n), r(n);
  for (long long int i = 0; i < n; i++) l[i] = r[i] = i;
  for (long long int i = 1; i < n; i++) {
    long long int st = i;
    while (st && a[i] >= a[st - 1]) {
      st = l[st - 1];
      l[i] = st;
    }
  }
  for (long long int i = n - 2; i >= 0; i--) {
    long long int st = i;
    while (st < n - 1 && a[i] > a[st + 1]) {
      st = r[st + 1];
      r[i] = st;
    }
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans += a[i] * (i - l[i] + 1) * (r[i] - i + 1);
  }
  cout << ans << "\n";
  return;
}
void work() {
  long long int n, q;
  cin >> n >> q;
  vector<long long int> a(N, 0ll);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  vector<long long int> b(n - 1);
  for (long long int i = 0; i + 1 < n; i++) b[i] = abs(a[i] - a[i + 1]);
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    x--, y--;
    result(y - x, vector<long long int>(b.begin() + x, b.begin() + y));
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    work();
  }
  return 0;
}

```