#include <bits/stdc++.h>
using namespace std;
const int nmax = 100005;
int n, q, x, y, a[nmax];
int lg[nmax], rmq[18][nmax], sol[18][nmax];
pair<int, int> get(int x, int y) {
  y--;
  int dif = y - x + 1;
  int mlc = lg[dif];
  if (rmq[mlc][x] > rmq[mlc][y - (1 << mlc) + 1])
    return make_pair(rmq[mlc][x], sol[mlc][x]);
  else
    return make_pair(rmq[mlc][y - (1 << mlc) + 1],
                     sol[mlc][y - (1 << mlc) + 1]);
}
long long solve(int x, int y) {
  if (x >= y) return 0;
  pair<int, int> mlc = get(x, y);
  int i = mlc.second;
  return solve(x, i) + solve(i + 1, y) +
         1LL * mlc.first * (i - x + 1) * (y - i);
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    rmq[0][i] = abs(a[i + 1] - a[i]);
    sol[0][i] = i;
  }
  for (int i = 2; i < n; i++) lg[i] = lg[i / 2] + 1;
  for (int i = 1, p = 2; p < n; i++, p *= 2)
    for (int j = 1; j < n; j++) {
      if (rmq[i - 1][j] > rmq[i - 1][j + p / 2]) {
        rmq[i][j] = rmq[i - 1][j];
        sol[i][j] = sol[i - 1][j];
      } else {
        rmq[i][j] = rmq[i - 1][j + p / 2];
        sol[i][j] = sol[i - 1][j + p / 2];
      }
    }
  for (; q; q--) {
    cin >> x >> y;
    cout << solve(x, y) << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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 nmax = 100005;
int n, q, x, y, a[nmax];
int lg[nmax], rmq[18][nmax], sol[18][nmax];
pair<int, int> get(int x, int y) {
  y--;
  int dif = y - x + 1;
  int mlc = lg[dif];
  if (rmq[mlc][x] > rmq[mlc][y - (1 << mlc) + 1])
    return make_pair(rmq[mlc][x], sol[mlc][x]);
  else
    return make_pair(rmq[mlc][y - (1 << mlc) + 1],
                     sol[mlc][y - (1 << mlc) + 1]);
}
long long solve(int x, int y) {
  if (x >= y) return 0;
  pair<int, int> mlc = get(x, y);
  int i = mlc.second;
  return solve(x, i) + solve(i + 1, y) +
         1LL * mlc.first * (i - x + 1) * (y - i);
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    rmq[0][i] = abs(a[i + 1] - a[i]);
    sol[0][i] = i;
  }
  for (int i = 2; i < n; i++) lg[i] = lg[i / 2] + 1;
  for (int i = 1, p = 2; p < n; i++, p *= 2)
    for (int j = 1; j < n; j++) {
      if (rmq[i - 1][j] > rmq[i - 1][j + p / 2]) {
        rmq[i][j] = rmq[i - 1][j];
        sol[i][j] = sol[i - 1][j];
      } else {
        rmq[i][j] = rmq[i - 1][j + p / 2];
        sol[i][j] = sol[i - 1][j + p / 2];
      }
    }
  for (; q; q--) {
    cin >> x >> y;
    cout << solve(x, y) << '\n';
  }
  return 0;
}

```