#include <bits/stdc++.h>
using namespace std;
int b[101010], a[101010], n, q;
int L[101010], R[101010];
int st[101010];
long long solve(int l, int r) {
  memset(b, 0, sizeof b);
  memset(L, 0, sizeof L);
  memset(R, 0, sizeof R);
  memset(st, 0, sizeof st);
  int m = 0;
  for (int i = l; i < r; i++) b[++m] = abs(a[i + 1] - a[i]);
  int sz = 0;
  st[sz] = 0;
  b[st[sz]] = 0;
  for (int i = 1; i <= m; i++) {
    while (sz && b[st[sz]] <= b[i]) --sz;
    L[i] = st[sz] + 1;
    st[++sz] = i;
  }
  sz = 0;
  st[sz] = m + 1;
  b[st[sz]] = 0;
  for (int i = m; i >= 1; i--) {
    while (sz && b[st[sz]] < b[i]) --sz;
    R[i] = st[sz] - 1;
    st[++sz] = i;
  }
  long long sum = 0ll;
  for (int i = 1; i <= n; i++) {
    sum += 1l * b[i] * (i - L[i] + 1ll) * (R[i] - i + 1ll);
  }
  return sum;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= q; i++) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r) << '\n';
  }
}

### 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;
int b[101010], a[101010], n, q;
int L[101010], R[101010];
int st[101010];
long long solve(int l, int r) {
  memset(b, 0, sizeof b);
  memset(L, 0, sizeof L);
  memset(R, 0, sizeof R);
  memset(st, 0, sizeof st);
  int m = 0;
  for (int i = l; i < r; i++) b[++m] = abs(a[i + 1] - a[i]);
  int sz = 0;
  st[sz] = 0;
  b[st[sz]] = 0;
  for (int i = 1; i <= m; i++) {
    while (sz && b[st[sz]] <= b[i]) --sz;
    L[i] = st[sz] + 1;
    st[++sz] = i;
  }
  sz = 0;
  st[sz] = m + 1;
  b[st[sz]] = 0;
  for (int i = m; i >= 1; i--) {
    while (sz && b[st[sz]] < b[i]) --sz;
    R[i] = st[sz] - 1;
    st[++sz] = i;
  }
  long long sum = 0ll;
  for (int i = 1; i <= n; i++) {
    sum += 1l * b[i] * (i - L[i] + 1ll) * (R[i] - i + 1ll);
  }
  return sum;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= q; i++) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r) << '\n';
  }
}

```