#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
const int N = 1e5 + 10;
const long long INF = 1LL << 60;
void solve() {
  long long n, q;
  cin >> n >> q;
  long long arr[n];
  long long d[n - 1];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    if (i != 0) d[i - 1] = abs(arr[i] - arr[i - 1]);
  }
  while (q--) {
    long long in[n], out[n];
    long long l, r;
    cin >> l >> r;
    l--, r--;
    stack<pair<long long, long long>> st;
    long long ans = 0;
    for (long long i = l; i < r; i++) {
      while (!st.empty() && st.top().first <= d[i]) {
        out[st.top().second] = i - 1;
        st.pop();
      }
      if (st.empty())
        in[i] = l;
      else
        in[i] = st.top().second + 1;
      st.push(make_pair(d[i], i));
    }
    while (!st.empty()) {
      out[st.top().second] = r - 1;
      st.pop();
    }
    for (long long i = l; i < r; i++) {
      ans += (i - in[i] + 1) * (out[i] - i + 1) * d[i];
    }
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long Tests = 1;
  while (Tests--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
const int N = 1e5 + 10;
const long long INF = 1LL << 60;
void solve() {
  long long n, q;
  cin >> n >> q;
  long long arr[n];
  long long d[n - 1];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    if (i != 0) d[i - 1] = abs(arr[i] - arr[i - 1]);
  }
  while (q--) {
    long long in[n], out[n];
    long long l, r;
    cin >> l >> r;
    l--, r--;
    stack<pair<long long, long long>> st;
    long long ans = 0;
    for (long long i = l; i < r; i++) {
      while (!st.empty() && st.top().first <= d[i]) {
        out[st.top().second] = i - 1;
        st.pop();
      }
      if (st.empty())
        in[i] = l;
      else
        in[i] = st.top().second + 1;
      st.push(make_pair(d[i], i));
    }
    while (!st.empty()) {
      out[st.top().second] = r - 1;
      st.pop();
    }
    for (long long i = l; i < r; i++) {
      ans += (i - in[i] + 1) * (out[i] - i + 1) * d[i];
    }
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long Tests = 1;
  while (Tests--) {
    solve();
  }
  return 0;
}

```