#include <bits/stdc++.h>
using namespace std;
void GO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
const int INF = (int)(1e9) + 1;
const int maxn = (int)2e5 + 1;
mt19937 rnd(time(0));
void solve() {
  int n, q;
  cin >> n >> q;
  vector<long long> a(n);
  for (auto &i : a) cin >> i;
  vector<long long> b(n - 1);
  n--;
  for (int i = 0; i < n; ++i) b[i] = abs(a[i] - a[i + 1]);
  vector<long long> l(n), r(n);
  vector<pair<long long, long long> > st;
  for (int i = 0; i < n; ++i) {
    while ((int)(st.size()) && b[i] >= st.back().first) st.pop_back();
    if (!(int)(st.size()))
      l[i] = 0;
    else
      l[i] = st.back().second + 1;
    st.push_back({b[i], i});
  }
  st.clear();
  for (int i = n - 1; i >= 0; --i) {
    while ((int)(st.size()) && b[i] > st.back().first) st.pop_back();
    if (!(int)(st.size()))
      r[i] = n - 1;
    else
      r[i] = st.back().second - 1;
    st.push_back({b[i], i});
  }
  while (q--) {
    long long lt, rt;
    cin >> lt >> rt;
    lt--, rt -= 2;
    long long ans = 0;
    for (int i = lt; i <= rt; ++i) {
      long long left = max(l[i], lt);
      long long right = min(r[i], rt);
      long long sum = i - left + right - i + 1;
      ans += (sum + (i - left) * (right - i)) * b[i];
    }
    cout << ans << '\n';
  }
}
signed main() {
  GO();
  int Q = 1;
  while (Q--) solve();
}

### Prompt

Create a solution in CPP for 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;
void GO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
const int INF = (int)(1e9) + 1;
const int maxn = (int)2e5 + 1;
mt19937 rnd(time(0));
void solve() {
  int n, q;
  cin >> n >> q;
  vector<long long> a(n);
  for (auto &i : a) cin >> i;
  vector<long long> b(n - 1);
  n--;
  for (int i = 0; i < n; ++i) b[i] = abs(a[i] - a[i + 1]);
  vector<long long> l(n), r(n);
  vector<pair<long long, long long> > st;
  for (int i = 0; i < n; ++i) {
    while ((int)(st.size()) && b[i] >= st.back().first) st.pop_back();
    if (!(int)(st.size()))
      l[i] = 0;
    else
      l[i] = st.back().second + 1;
    st.push_back({b[i], i});
  }
  st.clear();
  for (int i = n - 1; i >= 0; --i) {
    while ((int)(st.size()) && b[i] > st.back().first) st.pop_back();
    if (!(int)(st.size()))
      r[i] = n - 1;
    else
      r[i] = st.back().second - 1;
    st.push_back({b[i], i});
  }
  while (q--) {
    long long lt, rt;
    cin >> lt >> rt;
    lt--, rt -= 2;
    long long ans = 0;
    for (int i = lt; i <= rt; ++i) {
      long long left = max(l[i], lt);
      long long right = min(r[i], rt);
      long long sum = i - left + right - i + 1;
      ans += (sum + (i - left) * (right - i)) * b[i];
    }
    cout << ans << '\n';
  }
}
signed main() {
  GO();
  int Q = 1;
  while (Q--) solve();
}

```