#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 12;
int n, q, a[maxn], x[maxn], tmp[maxn];
void smax(int &a, int b) {
  if (b > a) a = b;
}
long long summax(int n, int a[]) {
  long long ans = 0;
  stack<pair<int, int> > s;
  s.push({-1, 1e9});
  for (int i = 0; i < n; i++) {
    while (s.top().second <= a[i]) s.pop();
    tmp[i] = i - s.top().first;
    s.push({i, a[i]});
  }
  s = stack<pair<int, int> >();
  s.push({n, 1e9});
  for (int i = n - 1; ~i; i--) {
    while (s.top().second < a[i]) s.pop();
    ans += a[i] * 1ll * tmp[i] * (s.top().first - i);
    s.push({i, a[i]});
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for_each(a, a + n, [](int &x) { cin >> x; });
  while (q--) {
    int l, r, ans = 0;
    cin >> l >> r;
    for (int i = l; i < r; i++) x[i - l] = abs(a[i] - a[i - 1]);
    printf("%lld\n", summax(r - l, x));
  }
  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 int maxn = 1e5 + 12;
int n, q, a[maxn], x[maxn], tmp[maxn];
void smax(int &a, int b) {
  if (b > a) a = b;
}
long long summax(int n, int a[]) {
  long long ans = 0;
  stack<pair<int, int> > s;
  s.push({-1, 1e9});
  for (int i = 0; i < n; i++) {
    while (s.top().second <= a[i]) s.pop();
    tmp[i] = i - s.top().first;
    s.push({i, a[i]});
  }
  s = stack<pair<int, int> >();
  s.push({n, 1e9});
  for (int i = n - 1; ~i; i--) {
    while (s.top().second < a[i]) s.pop();
    ans += a[i] * 1ll * tmp[i] * (s.top().first - i);
    s.push({i, a[i]});
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for_each(a, a + n, [](int &x) { cin >> x; });
  while (q--) {
    int l, r, ans = 0;
    cin >> l >> r;
    for (int i = l; i < r; i++) x[i - l] = abs(a[i] - a[i - 1]);
    printf("%lld\n", summax(r - l, x));
  }
  return 0;
}

```