#include <bits/stdc++.h>
using namespace std;
class unionfind {
 public:
  int par[100000];
  int ran[100000];
  int ren[100000];
  void init() {
    for (int i = 0; i < 100000; i++) {
      par[i] = i;
      ran[i] = 0;
      ren[i] = 1;
    }
  }
  int find(int a) {
    if (a == par[a])
      return a;
    else
      return par[a] = find(par[a]);
  }
  void unite(int a, int b) {
    a = find(a);
    b = find(b);
    if (a == b) return;
    if (ran[a] > ran[b]) {
      par[b] = a;
      ren[a] += ren[b];
    } else {
      par[a] = b;
      ren[b] += ren[a];
    }
    if (ran[a] == ran[b]) ran[b]++;
  }
};
unionfind uf;
int main() {
  int num, query;
  scanf("%d%d", &num, &query);
  vector<int> v;
  for (int i = 0; i < num; i++) {
    int z;
    scanf("%d", &z);
    v.push_back(z);
  }
  vector<pair<int, int> > dat;
  for (int i = 0; i < num - 1; i++) {
    dat.push_back(make_pair(abs(v[i + 1] - v[i]), i));
  }
  sort(dat.begin(), dat.end());
  for (int p = 0; p < query; p++) {
    long long ans = 0;
    uf.init();
    int za, zb;
    scanf("%d%d", &za, &zb);
    za--;
    zb--;
    for (int i = 0; i < dat.size(); i++) {
      if (za <= dat[i].second && dat[i].second + 1 <= zb) {
        ans += (long long)dat[i].first *
               (long long)uf.ren[uf.find(dat[i].second)] *
               (long long)uf.ren[uf.find(dat[i].second + 1)];
        uf.unite(dat[i].second, dat[i].second + 1);
      }
    }
    printf("%I64d\n", ans);
  }
}

### Prompt

In cpp, your task is to solve 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;
class unionfind {
 public:
  int par[100000];
  int ran[100000];
  int ren[100000];
  void init() {
    for (int i = 0; i < 100000; i++) {
      par[i] = i;
      ran[i] = 0;
      ren[i] = 1;
    }
  }
  int find(int a) {
    if (a == par[a])
      return a;
    else
      return par[a] = find(par[a]);
  }
  void unite(int a, int b) {
    a = find(a);
    b = find(b);
    if (a == b) return;
    if (ran[a] > ran[b]) {
      par[b] = a;
      ren[a] += ren[b];
    } else {
      par[a] = b;
      ren[b] += ren[a];
    }
    if (ran[a] == ran[b]) ran[b]++;
  }
};
unionfind uf;
int main() {
  int num, query;
  scanf("%d%d", &num, &query);
  vector<int> v;
  for (int i = 0; i < num; i++) {
    int z;
    scanf("%d", &z);
    v.push_back(z);
  }
  vector<pair<int, int> > dat;
  for (int i = 0; i < num - 1; i++) {
    dat.push_back(make_pair(abs(v[i + 1] - v[i]), i));
  }
  sort(dat.begin(), dat.end());
  for (int p = 0; p < query; p++) {
    long long ans = 0;
    uf.init();
    int za, zb;
    scanf("%d%d", &za, &zb);
    za--;
    zb--;
    for (int i = 0; i < dat.size(); i++) {
      if (za <= dat[i].second && dat[i].second + 1 <= zb) {
        ans += (long long)dat[i].first *
               (long long)uf.ren[uf.find(dat[i].second)] *
               (long long)uf.ren[uf.find(dat[i].second + 1)];
        uf.unite(dat[i].second, dat[i].second + 1);
      }
    }
    printf("%I64d\n", ans);
  }
}

```