#include <bits/stdc++.h>
using namespace std;
long long getRandomNumber() { return 4; }
struct pr {
  long long v;
  long long pos;
};
int a[110000];
int b[110000];
int lg[110000];
int st[30][110000];
int n, m;
void build2() {
  for (int i = 0; i <= n; i++)
    if (i > 1)
      lg[i] = lg[i / 2] + 1;
    else
      lg[i] = 0;
  for (int p = 0; (1 << p) < n; p++) {
    for (int i = 0; i + (1 << p) < n; i++) {
      if (p == 0)
        st[p][i] = i;
      else {
        if (b[st[p - 1][i]] > b[st[p - 1][i + (1 << (p - 1))]])
          st[p][i] = st[p - 1][i];
        else
          st[p][i] = st[p - 1][i + (1 << (p - 1))];
      }
    }
  }
}
long long get2(int l, int r) {
  int p = lg[r - l + 1];
  if (b[st[p][l]] > b[st[p][r - (1 << p) + 1]])
    return st[p][l];
  else
    return st[p][r - (1 << p) + 1];
}
int x1, x2;
long long sum = 0;
void rec(int l, int r) {
  pr ans;
  ans.pos = get2(l, r);
  ans.v = b[ans.pos];
  sum += 1ll * ans.v * (1ll * (ans.pos - l + 1) * (r - ans.pos + 1));
  if (ans.pos > l) rec(l, ans.pos - 1);
  if (ans.pos < r) rec(ans.pos + 1, r);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) b[i] = abs(a[i + 1] - a[i]);
  build2();
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x1, &x2);
    sum = 0;
    rec(x1 - 1, x2 - 2);
    cout << sum << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
long long getRandomNumber() { return 4; }
struct pr {
  long long v;
  long long pos;
};
int a[110000];
int b[110000];
int lg[110000];
int st[30][110000];
int n, m;
void build2() {
  for (int i = 0; i <= n; i++)
    if (i > 1)
      lg[i] = lg[i / 2] + 1;
    else
      lg[i] = 0;
  for (int p = 0; (1 << p) < n; p++) {
    for (int i = 0; i + (1 << p) < n; i++) {
      if (p == 0)
        st[p][i] = i;
      else {
        if (b[st[p - 1][i]] > b[st[p - 1][i + (1 << (p - 1))]])
          st[p][i] = st[p - 1][i];
        else
          st[p][i] = st[p - 1][i + (1 << (p - 1))];
      }
    }
  }
}
long long get2(int l, int r) {
  int p = lg[r - l + 1];
  if (b[st[p][l]] > b[st[p][r - (1 << p) + 1]])
    return st[p][l];
  else
    return st[p][r - (1 << p) + 1];
}
int x1, x2;
long long sum = 0;
void rec(int l, int r) {
  pr ans;
  ans.pos = get2(l, r);
  ans.v = b[ans.pos];
  sum += 1ll * ans.v * (1ll * (ans.pos - l + 1) * (r - ans.pos + 1));
  if (ans.pos > l) rec(l, ans.pos - 1);
  if (ans.pos < r) rec(ans.pos + 1, r);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) b[i] = abs(a[i + 1] - a[i]);
  build2();
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x1, &x2);
    sum = 0;
    rec(x1 - 1, x2 - 2);
    cout << sum << "\n";
  }
  return 0;
}

```