#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int IT_MAX;
const int MOD = 500500507;
const int INF = 1034567891;
const long long LL_INF = 2234567890123456789ll;
const double PI = 3.141592653589793238;
int in[100050];
int u[100050];
int l[100050];
int r[100050];
vector<pair<int, int> > V;
int main() {
  int N, Q, i, j, t1, t2;
  scanf("%d %d", &N, &Q);
  for (i = 1; i <= N; i++) scanf("%d", &u[i]);
  for (i = 1; i < N; i++) in[i] = -abs(u[i + 1] - u[i]);
  N--;
  l[1] = 1;
  V.push_back(pair<int, int>(in[1], 1));
  for (i = 2; i <= N; i++) {
    while (!V.empty() && V.back().first >= in[i]) V.pop_back();
    if (V.empty())
      l[i] = 1;
    else
      l[i] = V.back().second + 1;
    V.push_back(pair<int, int>(in[i], i));
  }
  V.clear();
  r[N] = N;
  V.push_back(pair<int, int>(in[N], N));
  for (i = N - 1; i >= 1; i--) {
    while (!V.empty() && V.back().first > in[i]) V.pop_back();
    if (V.empty())
      r[i] = N;
    else
      r[i] = V.back().second - 1;
    V.push_back(pair<int, int>(in[i], i));
  }
  while (Q--) {
    scanf("%d %d", &t1, &t2);
    t2--;
    long long ans = 0;
    for (i = t1; i <= t2; i++) {
      int x = max(t1, l[i]), y = min(t2, r[i]);
      ans += (long long)(i - x + 1) * (y - i + 1) * in[i];
    }
    printf("%lld\n", -ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#pragma warning(disable : 4996)
using namespace std;
int IT_MAX;
const int MOD = 500500507;
const int INF = 1034567891;
const long long LL_INF = 2234567890123456789ll;
const double PI = 3.141592653589793238;
int in[100050];
int u[100050];
int l[100050];
int r[100050];
vector<pair<int, int> > V;
int main() {
  int N, Q, i, j, t1, t2;
  scanf("%d %d", &N, &Q);
  for (i = 1; i <= N; i++) scanf("%d", &u[i]);
  for (i = 1; i < N; i++) in[i] = -abs(u[i + 1] - u[i]);
  N--;
  l[1] = 1;
  V.push_back(pair<int, int>(in[1], 1));
  for (i = 2; i <= N; i++) {
    while (!V.empty() && V.back().first >= in[i]) V.pop_back();
    if (V.empty())
      l[i] = 1;
    else
      l[i] = V.back().second + 1;
    V.push_back(pair<int, int>(in[i], i));
  }
  V.clear();
  r[N] = N;
  V.push_back(pair<int, int>(in[N], N));
  for (i = N - 1; i >= 1; i--) {
    while (!V.empty() && V.back().first > in[i]) V.pop_back();
    if (V.empty())
      r[i] = N;
    else
      r[i] = V.back().second - 1;
    V.push_back(pair<int, int>(in[i], i));
  }
  while (Q--) {
    scanf("%d %d", &t1, &t2);
    t2--;
    long long ans = 0;
    for (i = t1; i <= t2; i++) {
      int x = max(t1, l[i]), y = min(t2, r[i]);
      ans += (long long)(i - x + 1) * (y - i + 1) * in[i];
    }
    printf("%lld\n", -ans);
  }
  return 0;
}

```