#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const long long INFll = 1000000000000000000ll;
const int MOD = 1000000007;
const long double EPS = 0.0000000000001;
int n;
long long p[100500];
int a[100500];
int l[100500];
int r[100500];
int main() {
  ios_base::sync_with_stdio(0);
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  n--;
  for (int i = 0; i < n; ++i) {
    p[i] = abs(a[i + 1] - a[i]);
  }
  stack<pair<int, int> > t;
  for (int i = 0; i < n; ++i) {
    while (!t.empty() && t.top().first < p[i]) {
      t.pop();
    }
    l[i] = i + 1;
    if (t.size() > 0) {
      l[i] = i - t.top().second;
    }
    t.push(make_pair(p[i], i));
  }
  while (!t.empty()) {
    t.pop();
  }
  for (int i = n - 1; i >= 0; --i) {
    while (!t.empty() && t.top().first <= p[i]) {
      t.pop();
    }
    r[i] = n - i;
    if (t.size() > 0) {
      r[i] = t.top().second - i;
    }
    t.push(make_pair(p[i], i));
  }
  int L, R;
  long long a, b;
  long long ans = 0;
  for (int i = 0; i < q; ++i) {
    cin >> L >> R;
    L--;
    R -= 2;
    ans = 0;
    for (int i = L; i <= R; ++i) {
      a = l[i];
      b = r[i];
      if (i - l[i] + 1 < L) {
        a = i - L + 1;
      }
      if (i + r[i] - 1 > R) {
        b = R - i + 1;
      }
      ans += a * b * p[i];
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate 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 INF = 1000000000;
const long long INFll = 1000000000000000000ll;
const int MOD = 1000000007;
const long double EPS = 0.0000000000001;
int n;
long long p[100500];
int a[100500];
int l[100500];
int r[100500];
int main() {
  ios_base::sync_with_stdio(0);
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  n--;
  for (int i = 0; i < n; ++i) {
    p[i] = abs(a[i + 1] - a[i]);
  }
  stack<pair<int, int> > t;
  for (int i = 0; i < n; ++i) {
    while (!t.empty() && t.top().first < p[i]) {
      t.pop();
    }
    l[i] = i + 1;
    if (t.size() > 0) {
      l[i] = i - t.top().second;
    }
    t.push(make_pair(p[i], i));
  }
  while (!t.empty()) {
    t.pop();
  }
  for (int i = n - 1; i >= 0; --i) {
    while (!t.empty() && t.top().first <= p[i]) {
      t.pop();
    }
    r[i] = n - i;
    if (t.size() > 0) {
      r[i] = t.top().second - i;
    }
    t.push(make_pair(p[i], i));
  }
  int L, R;
  long long a, b;
  long long ans = 0;
  for (int i = 0; i < q; ++i) {
    cin >> L >> R;
    L--;
    R -= 2;
    ans = 0;
    for (int i = L; i <= R; ++i) {
      a = l[i];
      b = r[i];
      if (i - l[i] + 1 < L) {
        a = i - L + 1;
      }
      if (i + r[i] - 1 > R) {
        b = R - i + 1;
      }
      ans += a * b * p[i];
    }
    cout << ans << endl;
  }
  return 0;
}

```