#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n, q, ar[maxn], a[maxn], l[maxn], r[maxn], v[maxn], st[maxn];
long long calc(int x, int y) {
  if (y - x + 1 < 2) return 0;
  int m = y - x + 1;
  long long ret = 0;
  for (int i = 1; i <= m; i++) a[i] = ar[i + x - 1];
  for (int i = 1; i < m; i++) v[i] = abs(a[i] - a[i + 1]);
  int push = 0;
  for (int i = 1; i < m; i++) {
    while (push and v[st[push]] <= v[i]) push--;
    l[i] = st[push];
    st[++push] = i;
  }
  push = 0;
  for (int i = m - 1; i; i--) {
    while (push and v[st[push]] < v[i]) push--;
    if (!push)
      r[i] = m;
    else
      r[i] = st[push];
    st[++push] = i;
  }
  for (int i = 1; i < m; i++) ret += 1ll * (i - l[i]) * (r[i] - i) * v[i];
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]);
  for (int i = 1, l, r; i <= q; i++) {
    scanf("%d%d", &l, &r);
    cout << calc(l, r) << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 1e5 + 5;
int n, q, ar[maxn], a[maxn], l[maxn], r[maxn], v[maxn], st[maxn];
long long calc(int x, int y) {
  if (y - x + 1 < 2) return 0;
  int m = y - x + 1;
  long long ret = 0;
  for (int i = 1; i <= m; i++) a[i] = ar[i + x - 1];
  for (int i = 1; i < m; i++) v[i] = abs(a[i] - a[i + 1]);
  int push = 0;
  for (int i = 1; i < m; i++) {
    while (push and v[st[push]] <= v[i]) push--;
    l[i] = st[push];
    st[++push] = i;
  }
  push = 0;
  for (int i = m - 1; i; i--) {
    while (push and v[st[push]] < v[i]) push--;
    if (!push)
      r[i] = m;
    else
      r[i] = st[push];
    st[++push] = i;
  }
  for (int i = 1; i < m; i++) ret += 1ll * (i - l[i]) * (r[i] - i) * v[i];
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]);
  for (int i = 1, l, r; i <= q; i++) {
    scanf("%d%d", &l, &r);
    cout << calc(l, r) << '\n';
  }
  return 0;
}

```