#include <bits/stdc++.h>
const int inf = 1e9;
const int mod = (int)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
const int N = (int)1e5 + 10;
int n, a[N];
void solve() {
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<int> diff;
  for (int i = 0; i + 1 < n; ++i) {
    diff.push_back(abs(a[i] - a[i + 1]));
  };
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    vector<int> v;
    v.push_back(inf);
    for (int i = l; i + 1 <= r; ++i) {
      v.push_back(diff[i]);
    }
    stack<int> s;
    long long ans = 0;
    s.push(0);
    ;
    for (int i = 1; i < (int)(v.size()); ++i) {
      while (!s.empty() and v[i] > v[s.top()]) {
        int x = s.top();
        s.pop();
        int righthalf = i - x;
        int leftHalf = 1;
        if (!s.empty()) leftHalf = x - s.top();
        ans += v[x] * 1LL * righthalf * 1LL * leftHalf;
      }
      s.push(i);
    }
    while (!s.empty()) {
      int x = s.top();
      s.pop();
      if (x == 0) break;
      int leftHalf = 1;
      if (!s.empty()) leftHalf = x - s.top();
      ans += leftHalf * 1LL * v[x] * 1LL * ((int)(v.size()) - x);
    }
    cout << ans << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
    cout << '\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>
const int inf = 1e9;
const int mod = (int)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
const int N = (int)1e5 + 10;
int n, a[N];
void solve() {
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<int> diff;
  for (int i = 0; i + 1 < n; ++i) {
    diff.push_back(abs(a[i] - a[i + 1]));
  };
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    vector<int> v;
    v.push_back(inf);
    for (int i = l; i + 1 <= r; ++i) {
      v.push_back(diff[i]);
    }
    stack<int> s;
    long long ans = 0;
    s.push(0);
    ;
    for (int i = 1; i < (int)(v.size()); ++i) {
      while (!s.empty() and v[i] > v[s.top()]) {
        int x = s.top();
        s.pop();
        int righthalf = i - x;
        int leftHalf = 1;
        if (!s.empty()) leftHalf = x - s.top();
        ans += v[x] * 1LL * righthalf * 1LL * leftHalf;
      }
      s.push(i);
    }
    while (!s.empty()) {
      int x = s.top();
      s.pop();
      if (x == 0) break;
      int leftHalf = 1;
      if (!s.empty()) leftHalf = x - s.top();
      ans += leftHalf * 1LL * v[x] * 1LL * ((int)(v.size()) - x);
    }
    cout << ans << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```