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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  set<int> done;
  vector<pair<int, int> > grads1, grads2;
  vector<long long> sols;
  int n, q;
  scanf("%d%d", &n, &q);
  vector<int> vals;
  vals.resize(n);
  for (int c = 0; c < n; c++) scanf("%d", &vals[c]);
  for (int c = 0; c < n - 1; c++)
    grads1.push_back(pair<int, int>(abs(vals[c + 1] - vals[c]), c));
  grads2 = grads1;
  sort(grads2.begin(), grads2.end());
  sols.resize(n - 1);
  vector<int> left(n - 1), right(n - 1);
  for (int c = n - 2; c >= 0; c--) {
    auto l = done.upper_bound(grads2[c].second);
    auto u = done.lower_bound(grads2[c].second + 1);
    if (l == done.begin())
      left[grads2[c].second] = 0;
    else
      left[grads2[c].second] = *(--l);
    if (u == done.end())
      right[grads2[c].second] = n - 1;
    else
      right[grads2[c].second] = *u;
    done.insert(grads2[c].second);
    done.insert(grads2[c].second + 1);
  }
  int i, j;
  while (q--) {
    scanf("%d%d", &i, &j);
    long long r = 0;
    i--;
    j--;
    for (int c = i; c < j; c++) {
      r += (long long)grads1[c].first * (min(right[c], j) - c) *
           (c + 1 - max(left[c], i));
    }
    cout << r << endl;
  }
  return 0;
}
