#include <bits/stdc++.h>
using namespace std;
long long MOD = 1000000007;
int inp[100100];
void solve(int a[], int n) {
  stack<std::pair<int, int> > st;
  long long ans = 0;
  st.push(make_pair(a[0], 0));
  int pres = a[n];
  a[n] = MOD;
  for (int i = 1; i <= n; ++i) {
    if (st.empty() || st.top().first >= a[i]) {
      st.push(make_pair(a[i], i));
      continue;
    }
    std::pair<int, int> p = st.top();
    st.pop();
    int l = 0;
    if (!st.empty()) l = st.top().second + 1;
    ans += (p.second - l + 1) * 1LL * (i - p.second) * 1LL * p.first;
    i -= 1;
  }
  a[n] = pres;
  printf("%lld\n", ans);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < (n); ++i) scanf("%d", inp + i);
  for (int i = n - 1; i > 0; --i) {
    inp[i] = abs(inp[i] - inp[i - 1]);
  }
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l, --r;
    solve(inp + l + 1, r - l);
  }
  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;
long long MOD = 1000000007;
int inp[100100];
void solve(int a[], int n) {
  stack<std::pair<int, int> > st;
  long long ans = 0;
  st.push(make_pair(a[0], 0));
  int pres = a[n];
  a[n] = MOD;
  for (int i = 1; i <= n; ++i) {
    if (st.empty() || st.top().first >= a[i]) {
      st.push(make_pair(a[i], i));
      continue;
    }
    std::pair<int, int> p = st.top();
    st.pop();
    int l = 0;
    if (!st.empty()) l = st.top().second + 1;
    ans += (p.second - l + 1) * 1LL * (i - p.second) * 1LL * p.first;
    i -= 1;
  }
  a[n] = pres;
  printf("%lld\n", ans);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < (n); ++i) scanf("%d", inp + i);
  for (int i = n - 1; i > 0; --i) {
    inp[i] = abs(inp[i] - inp[i - 1]);
  }
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l, --r;
    solve(inp + l + 1, r - l);
  }
  return 0;
}

```