#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
const int INF = 1e9;
int a[N], b[N], num, l[N], r[N], f[N], n, m, c[N];
map<int, int> mm;
int lowbit(int x) { return x & (-x); }
int get(int x) {
  x = num - x;
  if (x <= 0) return 0;
  int s = -INF;
  for (int i = x; i; i -= lowbit(i)) s = max(s, f[i]);
  return s;
}
void add(int x, int t) {
  x = num - x;
  if (x <= 0) return;
  for (int i = x; i <= n; i += lowbit(i)) f[i] = max(f[i], t);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n - 1; i++) {
    b[i] = abs(a[i] - a[i + 1]);
    c[i] = b[i];
    mm[b[i]] = 1;
  }
  num = 0;
  for (map<int, int>::iterator it = mm.begin(); it != mm.end(); it++)
    (*it).second = ++num;
  num++;
  for (int i = 1; i < n; i++) b[i] = mm[b[i]];
  for (int i = 1; i < n; i++) {
    l[i] = get(b[i] + 1);
    add(b[i], i);
  }
  for (int i = 1; i <= n; i++) f[i] = -n;
  for (int i = n - 1; i >= 1; i--) {
    r[i] = -get(b[i]);
    add(b[i], -i);
  }
  for (int i = 1; i <= m; i++) {
    int p, q;
    scanf("%d%d", &p, &q);
    long long ans = 0;
    for (int i = p; i < q; i++) {
      long long t1 = max(p - 1, l[i]);
      long long t2 = min(q, r[i]);
      ans += (long long)(t2 - i) * (long long)(i - t1) * (long long)c[i];
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const int N = 2e5;
const int INF = 1e9;
int a[N], b[N], num, l[N], r[N], f[N], n, m, c[N];
map<int, int> mm;
int lowbit(int x) { return x & (-x); }
int get(int x) {
  x = num - x;
  if (x <= 0) return 0;
  int s = -INF;
  for (int i = x; i; i -= lowbit(i)) s = max(s, f[i]);
  return s;
}
void add(int x, int t) {
  x = num - x;
  if (x <= 0) return;
  for (int i = x; i <= n; i += lowbit(i)) f[i] = max(f[i], t);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n - 1; i++) {
    b[i] = abs(a[i] - a[i + 1]);
    c[i] = b[i];
    mm[b[i]] = 1;
  }
  num = 0;
  for (map<int, int>::iterator it = mm.begin(); it != mm.end(); it++)
    (*it).second = ++num;
  num++;
  for (int i = 1; i < n; i++) b[i] = mm[b[i]];
  for (int i = 1; i < n; i++) {
    l[i] = get(b[i] + 1);
    add(b[i], i);
  }
  for (int i = 1; i <= n; i++) f[i] = -n;
  for (int i = n - 1; i >= 1; i--) {
    r[i] = -get(b[i]);
    add(b[i], -i);
  }
  for (int i = 1; i <= m; i++) {
    int p, q;
    scanf("%d%d", &p, &q);
    long long ans = 0;
    for (int i = p; i < q; i++) {
      long long t1 = max(p - 1, l[i]);
      long long t2 = min(q, r[i]);
      ans += (long long)(t2 - i) * (long long)(i - t1) * (long long)c[i];
    }
    cout << ans << endl;
  }
  return 0;
}

```