#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (1 << 20);
long long n, q;
long long a[MAXN];
void read() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
}
long long next_bad_word[MAXN], prev_bad_word[MAXN];
long long solve(long long lbound, long long rbound) {
  long long ans = 0;
  for (long long i = lbound; i <= rbound - 1; i++) {
    long long l = max(prev_bad_word[i] + 1, lbound),
              r = min(next_bad_word[i] - 1, rbound - 1);
    ans += a[i] * (i - l + 1) * (r - i + 1);
  }
  return ans;
}
long long abs_my_function(long long v) { return v < 0 ? -v : v; }
void solve() {
  for (long long i = 1; i <= n - 1; i++)
    a[i] = abs_my_function(a[i] - a[i + 1]);
  stack<long long> st;
  for (long long i = 1; i <= n - 1; i++) {
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    if (st.empty())
      prev_bad_word[i] = 0;
    else
      prev_bad_word[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (long long i = n - 1; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      next_bad_word[i] = n;
    else
      next_bad_word[i] = st.top();
    st.push(i);
  }
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(l, r) << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  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 long long MAXN = (1 << 20);
long long n, q;
long long a[MAXN];
void read() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
}
long long next_bad_word[MAXN], prev_bad_word[MAXN];
long long solve(long long lbound, long long rbound) {
  long long ans = 0;
  for (long long i = lbound; i <= rbound - 1; i++) {
    long long l = max(prev_bad_word[i] + 1, lbound),
              r = min(next_bad_word[i] - 1, rbound - 1);
    ans += a[i] * (i - l + 1) * (r - i + 1);
  }
  return ans;
}
long long abs_my_function(long long v) { return v < 0 ? -v : v; }
void solve() {
  for (long long i = 1; i <= n - 1; i++)
    a[i] = abs_my_function(a[i] - a[i + 1]);
  stack<long long> st;
  for (long long i = 1; i <= n - 1; i++) {
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    if (st.empty())
      prev_bad_word[i] = 0;
    else
      prev_bad_word[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (long long i = n - 1; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      next_bad_word[i] = n;
    else
      next_bad_word[i] = st.top();
    st.push(i);
  }
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(l, r) << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}

```