#include <bits/stdc++.h>
using namespace std;
int N, Q;
int a[100010];
int ql[110], qr[110];
long long ans[110];
void query(int L, int M, int R, long long x) {
  int i;
  for ((i) = 0; (i) < (int)(Q); (i)++)
    if (ql[i] < M && qr[i] > M) {
      long long tmp1 = M - max(L, ql[i]);
      long long tmp2 = min(R, qr[i]) - M;
      ans[i] += tmp1 * tmp2 * x;
    }
}
int main(void) {
  int i, j;
  cin >> N >> Q;
  for ((i) = 0; (i) < (int)(N); (i)++) scanf("%d", &a[i]);
  for ((i) = 0; (i) < (int)(Q); (i)++) {
    scanf("%d%d", &ql[i], &qr[i]);
    ql[i]--;
  }
  set<int> s;
  s.insert(0);
  s.insert(N);
  vector<pair<int, int> > v;
  for (i = 1; i < N; i++) {
    int tmp = a[i] - a[i - 1];
    if (tmp < 0) tmp = -tmp;
    v.push_back(make_pair(tmp, i));
  }
  sort(v.begin(), v.end());
  int sz = v.size();
  for (i = sz - 1; i >= 0; i--) {
    long long d = v[i].first;
    int id = v[i].second;
    s.insert(id);
    __typeof(s.begin()) itr = s.find(id);
    itr--;
    int L = *(itr);
    itr++;
    itr++;
    int R = *(itr);
    query(L, id, R, d);
  }
  for ((i) = 0; (i) < (int)(Q); (i)++) cout << ans[i] << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int N, Q;
int a[100010];
int ql[110], qr[110];
long long ans[110];
void query(int L, int M, int R, long long x) {
  int i;
  for ((i) = 0; (i) < (int)(Q); (i)++)
    if (ql[i] < M && qr[i] > M) {
      long long tmp1 = M - max(L, ql[i]);
      long long tmp2 = min(R, qr[i]) - M;
      ans[i] += tmp1 * tmp2 * x;
    }
}
int main(void) {
  int i, j;
  cin >> N >> Q;
  for ((i) = 0; (i) < (int)(N); (i)++) scanf("%d", &a[i]);
  for ((i) = 0; (i) < (int)(Q); (i)++) {
    scanf("%d%d", &ql[i], &qr[i]);
    ql[i]--;
  }
  set<int> s;
  s.insert(0);
  s.insert(N);
  vector<pair<int, int> > v;
  for (i = 1; i < N; i++) {
    int tmp = a[i] - a[i - 1];
    if (tmp < 0) tmp = -tmp;
    v.push_back(make_pair(tmp, i));
  }
  sort(v.begin(), v.end());
  int sz = v.size();
  for (i = sz - 1; i >= 0; i--) {
    long long d = v[i].first;
    int id = v[i].second;
    s.insert(id);
    __typeof(s.begin()) itr = s.find(id);
    itr--;
    int L = *(itr);
    itr++;
    itr++;
    int R = *(itr);
    query(L, id, R, d);
  }
  for ((i) = 0; (i) < (int)(Q); (i)++) cout << ans[i] << endl;
  return 0;
}

```