#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int res = 0;
  int neg;
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9' || ch == '-') {
      if (ch == '-')
        neg = -1;
      else
        neg = 1, res = ch - '0';
      break;
    }
  }
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9')
      res *= 10, res += ch - '0';
    else
      break;
  }
  return res * neg;
}
const int maxn = 100020;
const int maxm = 1000020;
const int MOd = 1e9 + 7;
const int P = 37;
struct node {
  int l, r, maxi;
};
node segment[1 << 18];
int a, q, ar[maxn];
int n;
pair<long long, int> st[maxn];
int at;
int main() {
  scanf("%d %d", &a, &q);
  for (int i = 1; i <= a; i++) scanf("%d", &ar[i]);
  for (int i = 1; i <= a; i++) ar[i] = abs(ar[i] - ar[i + 1]);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    long long t = 0, ans = 0;
    at = 0;
    st[++at] = pair<long long, int>(0, l - 1);
    for (int i = l; i < r; i++) {
      while (at > 1 && st[at].first <= ar[i]) {
        t -= (long long)(st[at].second - st[at - 1].second) * st[at].first;
        at--;
      }
      t += (long long)(i - st[at].second) * ar[i];
      st[++at] = pair<long long, int>(ar[i], i);
      ans += t;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
inline int read() {
  int res = 0;
  int neg;
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9' || ch == '-') {
      if (ch == '-')
        neg = -1;
      else
        neg = 1, res = ch - '0';
      break;
    }
  }
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9')
      res *= 10, res += ch - '0';
    else
      break;
  }
  return res * neg;
}
const int maxn = 100020;
const int maxm = 1000020;
const int MOd = 1e9 + 7;
const int P = 37;
struct node {
  int l, r, maxi;
};
node segment[1 << 18];
int a, q, ar[maxn];
int n;
pair<long long, int> st[maxn];
int at;
int main() {
  scanf("%d %d", &a, &q);
  for (int i = 1; i <= a; i++) scanf("%d", &ar[i]);
  for (int i = 1; i <= a; i++) ar[i] = abs(ar[i] - ar[i + 1]);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    long long t = 0, ans = 0;
    at = 0;
    st[++at] = pair<long long, int>(0, l - 1);
    for (int i = l; i < r; i++) {
      while (at > 1 && st[at].first <= ar[i]) {
        t -= (long long)(st[at].second - st[at - 1].second) * st[at].first;
        at--;
      }
      t += (long long)(i - st[at].second) * ar[i];
      st[++at] = pair<long long, int>(ar[i], i);
      ans += t;
    }
    cout << ans << endl;
  }
  return 0;
}

```