#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e5 + 123;
const int INF = (int)1e9 + 123;
const double eps = 1e-6;
long long n, q, a[N], l, r, ans;
stack<pair<long long, pair<long long, long long> > > s;
long long l_border[N], r_border[N];
long long solve(long long lef, long long rig) {
  ans = 0;
  if (r - l + 1 < 2) {
    return 0;
  }
  vector<long long> now;
  for (int i = l + 1; i <= r; i++) {
    now.push_back(abs(a[i] - a[i - 1]));
  }
  s.push(make_pair(now[0], make_pair(0ll, 0ll)));
  for (int i = 1; i < now.size(); i++) {
    l_border[i] = i;
    while (s.size() && now[i] >= s.top().first) {
      l_border[i] = l_border[s.top().second.second];
      s.pop();
    }
    s.push(make_pair(now[i], make_pair(l_border[i], i)));
  }
  while (s.size()) {
    s.pop();
  }
  r_border[now.size() - 1] = now.size() - 1;
  s.push(make_pair(now[now.size() - 1],
                   make_pair(now.size() - 1ll, now.size() - 1ll)));
  for (int i = now.size() - 2; i >= 0; i--) {
    r_border[i] = i;
    while (s.size() && now[i] > s.top().first) {
      r_border[i] = r_border[s.top().second.second];
      s.pop();
    }
    s.push(make_pair(now[i], make_pair(r_border[i], i)));
  }
  while (s.size()) {
    s.pop();
  }
  for (int i = 0; i < now.size(); i++) {
    ans += (r_border[i] - i + i - l_border[i] +
            (i - l_border[i]) * (r_border[i] - i)) *
               now[i] +
           now[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    cout << solve(l, r) << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int N = (int)1e5 + 123;
const int INF = (int)1e9 + 123;
const double eps = 1e-6;
long long n, q, a[N], l, r, ans;
stack<pair<long long, pair<long long, long long> > > s;
long long l_border[N], r_border[N];
long long solve(long long lef, long long rig) {
  ans = 0;
  if (r - l + 1 < 2) {
    return 0;
  }
  vector<long long> now;
  for (int i = l + 1; i <= r; i++) {
    now.push_back(abs(a[i] - a[i - 1]));
  }
  s.push(make_pair(now[0], make_pair(0ll, 0ll)));
  for (int i = 1; i < now.size(); i++) {
    l_border[i] = i;
    while (s.size() && now[i] >= s.top().first) {
      l_border[i] = l_border[s.top().second.second];
      s.pop();
    }
    s.push(make_pair(now[i], make_pair(l_border[i], i)));
  }
  while (s.size()) {
    s.pop();
  }
  r_border[now.size() - 1] = now.size() - 1;
  s.push(make_pair(now[now.size() - 1],
                   make_pair(now.size() - 1ll, now.size() - 1ll)));
  for (int i = now.size() - 2; i >= 0; i--) {
    r_border[i] = i;
    while (s.size() && now[i] > s.top().first) {
      r_border[i] = r_border[s.top().second.second];
      s.pop();
    }
    s.push(make_pair(now[i], make_pair(r_border[i], i)));
  }
  while (s.size()) {
    s.pop();
  }
  for (int i = 0; i < now.size(); i++) {
    ans += (r_border[i] - i + i - l_border[i] +
            (i - l_border[i]) * (r_border[i] - i)) *
               now[i] +
           now[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    cout << solve(l, r) << "\n";
  }
  return 0;
}

```