#include <bits/stdc++.h>
using namespace std;
int n, n_, k, q;
vector<int> a, b;
long long res;
vector<vector<int> > spt, ind;
void build_spt() {
  spt.assign(k, vector<int>(n, 0));
  ind.assign(k, vector<int>(n, 0));
  for (int j = 0; j < n; ++j) {
    spt[0][j] = b[j];
    ind[0][j] = j;
  }
  int m = 1;
  for (int i = 1; i < k; ++i) {
    m <<= 1;
    for (int j = 0; j < n - m + 1; ++j) {
      if (spt[i - 1][j] > spt[i - 1][j + m / 2]) {
        spt[i][j] = spt[i - 1][j];
        ind[i][j] = ind[i - 1][j];
      } else {
        spt[i][j] = spt[i - 1][j + m / 2];
        ind[i][j] = ind[i - 1][j + m / 2];
      }
    }
  }
}
void calc_n() {
  n = 1;
  k = 1;
  while (n < n_) {
    n <<= 1;
    ++k;
  }
}
void solve(int l, int r) {
  if (l == r) {
    res += (long long)b[l];
    return;
  }
  int len = r - l + 1;
  int k = 0;
  int m = 1;
  while (m + m < len) {
    m <<= 1;
    ++k;
  }
  long long mb;
  int in;
  if (spt[k][l] > spt[k][r - m + 1]) {
    mb = spt[k][l];
    in = ind[k][l];
  } else {
    mb = spt[k][r - m + 1];
    in = ind[k][r - m + 1];
  }
  res += (mb * ((r - in + 1ll) * 1ll * (in - l + 1ll)));
  if (in > l) solve(l, in - 1);
  if (in < r) solve(in + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n_ >> q;
  calc_n();
  a.assign(n + 1, 0);
  for (int i = 0; i < n_; ++i) cin >> a[i];
  b.resize(n);
  for (int i = 0; i < n; ++i) b[i] = abs(a[i + 1] - a[i]);
  build_spt();
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    --r;
    res = 0;
    solve(l, r);
    cout << res << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n, n_, k, q;
vector<int> a, b;
long long res;
vector<vector<int> > spt, ind;
void build_spt() {
  spt.assign(k, vector<int>(n, 0));
  ind.assign(k, vector<int>(n, 0));
  for (int j = 0; j < n; ++j) {
    spt[0][j] = b[j];
    ind[0][j] = j;
  }
  int m = 1;
  for (int i = 1; i < k; ++i) {
    m <<= 1;
    for (int j = 0; j < n - m + 1; ++j) {
      if (spt[i - 1][j] > spt[i - 1][j + m / 2]) {
        spt[i][j] = spt[i - 1][j];
        ind[i][j] = ind[i - 1][j];
      } else {
        spt[i][j] = spt[i - 1][j + m / 2];
        ind[i][j] = ind[i - 1][j + m / 2];
      }
    }
  }
}
void calc_n() {
  n = 1;
  k = 1;
  while (n < n_) {
    n <<= 1;
    ++k;
  }
}
void solve(int l, int r) {
  if (l == r) {
    res += (long long)b[l];
    return;
  }
  int len = r - l + 1;
  int k = 0;
  int m = 1;
  while (m + m < len) {
    m <<= 1;
    ++k;
  }
  long long mb;
  int in;
  if (spt[k][l] > spt[k][r - m + 1]) {
    mb = spt[k][l];
    in = ind[k][l];
  } else {
    mb = spt[k][r - m + 1];
    in = ind[k][r - m + 1];
  }
  res += (mb * ((r - in + 1ll) * 1ll * (in - l + 1ll)));
  if (in > l) solve(l, in - 1);
  if (in < r) solve(in + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n_ >> q;
  calc_n();
  a.assign(n + 1, 0);
  for (int i = 0; i < n_; ++i) cin >> a[i];
  b.resize(n);
  for (int i = 0; i < n; ++i) b[i] = abs(a[i + 1] - a[i]);
  build_spt();
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    --r;
    res = 0;
    solve(l, r);
    cout << res << endl;
  }
  return 0;
}

```