#include <bits/stdc++.h>
using namespace ::std;
const int N = 100000;
int n, m;
long long a[N + 5];
int uf[N + 5];
long long s[N + 5];
int up(int now) {
  if (uf[now] == now) return now;
  return uf[now] = up(uf[now]);
}
void uf_merge(int v1, int v2) {
  int w1 = up(v1), w2 = up(v2);
  if (w1 != w2) {
    uf[w1] = w2;
    s[w2] += s[w1];
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < m; i++) {
    int l, r;
    long long ans = 0;
    scanf("%d %d", &l, &r);
    for (int j = l; j <= r; j++) uf[j] = j, s[j] = 1;
    vector<pair<long long, int> > v;
    for (int j = l; j < r; j++)
      v.push_back(pair<long long, int>(abs(a[j + 1] - a[j]), j));
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); i++) {
      int v1 = up(v[i].second);
      int v2 = up(v[i].second + 1);
      ans += v[i].first * s[v1] * s[v2];
      uf_merge(v[i].second, v[i].second + 1);
    }
    printf("%I64d\n", ans);
  }
  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 N = 100000;
int n, m;
long long a[N + 5];
int uf[N + 5];
long long s[N + 5];
int up(int now) {
  if (uf[now] == now) return now;
  return uf[now] = up(uf[now]);
}
void uf_merge(int v1, int v2) {
  int w1 = up(v1), w2 = up(v2);
  if (w1 != w2) {
    uf[w1] = w2;
    s[w2] += s[w1];
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < m; i++) {
    int l, r;
    long long ans = 0;
    scanf("%d %d", &l, &r);
    for (int j = l; j <= r; j++) uf[j] = j, s[j] = 1;
    vector<pair<long long, int> > v;
    for (int j = l; j < r; j++)
      v.push_back(pair<long long, int>(abs(a[j + 1] - a[j]), j));
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); i++) {
      int v1 = up(v[i].second);
      int v2 = up(v[i].second + 1);
      ans += v[i].first * s[v1] * s[v2];
      uf_merge(v[i].second, v[i].second + 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```