#include <bits/stdc++.h>
using namespace std;
template <typename C>
void maxi(C& a, C b) {
  if (a < b) a = b;
}
template <typename C>
void mini(C& a, C b) {
  if (a > b) a = b;
}
template <class TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char* sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
int n, q;
int t[100010];
void test() {
  int a, b;
  scanf("%d%d", &a, &b);
  ("test", a, b);
  a--;
  b--;
  long long war = 0;
  long long res = 0;
  vector<pair<int, int> > stos;
  for (int i = a; i < b; i++) {
    (t[i]);
    int sze = 1;
    while (((int)(stos).size()) > 0 && stos.back().first < t[i]) {
      war -= stos.back().first * 1ll * stos.back().second;
      sze += stos.back().second;
      stos.pop_back();
      ("tu");
    }
    (t[i], sze);
    stos.push_back({t[i], sze});
    war += t[i] * 1ll * sze;
    res += war;
  }
  printf("%lld\n", res);
}
int main() {
  scanf("%d%d", &n, &q);
  int ost;
  scanf("%d", &ost);
  for (int i = 0; i < n - 1; i++) {
    int pom;
    scanf("%d", &pom);
    t[i] = abs(pom - ost);
    ost = pom;
  }
  while (q--) test();
}

### 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;
template <typename C>
void maxi(C& a, C b) {
  if (a < b) a = b;
}
template <typename C>
void mini(C& a, C b) {
  if (a > b) a = b;
}
template <class TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char* sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
int n, q;
int t[100010];
void test() {
  int a, b;
  scanf("%d%d", &a, &b);
  ("test", a, b);
  a--;
  b--;
  long long war = 0;
  long long res = 0;
  vector<pair<int, int> > stos;
  for (int i = a; i < b; i++) {
    (t[i]);
    int sze = 1;
    while (((int)(stos).size()) > 0 && stos.back().first < t[i]) {
      war -= stos.back().first * 1ll * stos.back().second;
      sze += stos.back().second;
      stos.pop_back();
      ("tu");
    }
    (t[i], sze);
    stos.push_back({t[i], sze});
    war += t[i] * 1ll * sze;
    res += war;
  }
  printf("%lld\n", res);
}
int main() {
  scanf("%d%d", &n, &q);
  int ost;
  scanf("%d", &ost);
  for (int i = 0; i < n - 1; i++) {
    int pom;
    scanf("%d", &pom);
    t[i] = abs(pom - ost);
    ost = pom;
  }
  while (q--) test();
}

```