#include <bits/stdc++.h>
using namespace std;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
  }
} _;
const int BOUND = 100001;
long long input[BOUND];
long long d[BOUND];
long long a[BOUND];
long long b[BOUND];
template <typename T>
void print(T first, T last) {
  for (auto it = first; it != last; ++it) {
    cout << *it << " ";
  }
  cout << endl;
}
long long solve(long long l, long long r) {
  long long n = r - l;
  fill(a + l, a + r, l - 1);
  fill(b + l, b + r, r);
  stack<pair<long long, long long> > gner;
  gner.push(make_pair(d[r - 1], r - 1));
  for (long long i = 1; i < n; ++i) {
    long long next = d[r - 1 - i];
    while (not gner.empty() and gner.top().first <= next) {
      pair<long long, long long> found = gner.top();
      gner.pop();
      a[found.second] = r - 1 - i;
    }
    gner.push(make_pair(next, r - 1 - i));
  }
  stack<pair<long long, long long> > gne;
  gne.push(make_pair(d[l], l));
  for (long long i = 1; i < n; ++i) {
    long long next = d[l + i];
    while (not gne.empty() and gne.top().first < next) {
      pair<long long, long long> found = gne.top();
      gne.pop();
      b[found.second] = l + i;
    }
    gne.push(make_pair(next, l + i));
  }
  long long result = 0;
  for (long long i = l; i < r; ++i) {
    long long p = (i - a[i]) * (b[i] - i);
    result += d[i] * p;
  }
  return result;
}
int main() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) {
    int t;
    cin >> t;
    input[i] = t;
  }
  for (long long i = 1; i <= n - 1; ++i) {
    long long tmp = input[i + 1] - input[i];
    d[i] = (tmp < 0 ? -tmp : tmp);
  }
  for (int i = 0; i < q; ++i) {
    long long l, r;
    cin >> l >> r;
    long long result = solve(l, r);
    cout << result << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
  }
} _;
const int BOUND = 100001;
long long input[BOUND];
long long d[BOUND];
long long a[BOUND];
long long b[BOUND];
template <typename T>
void print(T first, T last) {
  for (auto it = first; it != last; ++it) {
    cout << *it << " ";
  }
  cout << endl;
}
long long solve(long long l, long long r) {
  long long n = r - l;
  fill(a + l, a + r, l - 1);
  fill(b + l, b + r, r);
  stack<pair<long long, long long> > gner;
  gner.push(make_pair(d[r - 1], r - 1));
  for (long long i = 1; i < n; ++i) {
    long long next = d[r - 1 - i];
    while (not gner.empty() and gner.top().first <= next) {
      pair<long long, long long> found = gner.top();
      gner.pop();
      a[found.second] = r - 1 - i;
    }
    gner.push(make_pair(next, r - 1 - i));
  }
  stack<pair<long long, long long> > gne;
  gne.push(make_pair(d[l], l));
  for (long long i = 1; i < n; ++i) {
    long long next = d[l + i];
    while (not gne.empty() and gne.top().first < next) {
      pair<long long, long long> found = gne.top();
      gne.pop();
      b[found.second] = l + i;
    }
    gne.push(make_pair(next, l + i));
  }
  long long result = 0;
  for (long long i = l; i < r; ++i) {
    long long p = (i - a[i]) * (b[i] - i);
    result += d[i] * p;
  }
  return result;
}
int main() {
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) {
    int t;
    cin >> t;
    input[i] = t;
  }
  for (long long i = 1; i <= n - 1; ++i) {
    long long tmp = input[i + 1] - input[i];
    d[i] = (tmp < 0 ? -tmp : tmp);
  }
  for (int i = 0; i < q; ++i) {
    long long l, r;
    cin >> l >> r;
    long long result = solve(l, r);
    cout << result << endl;
  }
  return 0;
}

```