#include <bits/stdc++.h>
using namespace std;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
long long n, q, a[100005], b[100005];
pair<long long, long long> p[100005];
long long ans(long long l, long long r) {
  long long ans = 0;
  long long sum = 0;
  long long cur = 0;
  for (long long i = l; i < r; i++) {
    long long range = 1;
    while (cur && p[cur].first <= b[i]) {
      sum -= (p[cur].first * p[cur].second);
      range += p[cur].second;
      cur--;
    }
    p[++cur].first = b[i];
    p[cur].second = range;
    sum += range * b[i];
    ans += sum;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i < n; i++) b[i] = abs(a[i] - a[i + 1]);
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << ans(l, r) << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
long long n, q, a[100005], b[100005];
pair<long long, long long> p[100005];
long long ans(long long l, long long r) {
  long long ans = 0;
  long long sum = 0;
  long long cur = 0;
  for (long long i = l; i < r; i++) {
    long long range = 1;
    while (cur && p[cur].first <= b[i]) {
      sum -= (p[cur].first * p[cur].second);
      range += p[cur].second;
      cur--;
    }
    p[++cur].first = b[i];
    p[cur].second = range;
    sum += range * b[i];
    ans += sum;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i < n; i++) b[i] = abs(a[i] - a[i + 1]);
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << ans(l, r) << endl;
  }
  return 0;
}

```