#include <bits/stdc++.h>
using namespace std;
const long long maxN = 1e6 + 5, MOD = 1e9 + 7, INF = 1e9;
const double PI = acos(-1), eps = 1e-7;
int n, q, A[maxN], D[maxN];
long long r[maxN], l[maxN], st[maxN];
int main() {
  cin >> n >> q;
  for (int i = int(1); i < int(n + 1); i++) {
    cin >> A[i];
  }
  A[0] = A[n + 1] = INF;
  for (int i = int(1); i < int(n + 1); i++) {
    D[i - 1] = abs(A[i] - A[i - 1]);
  }
  D[n] = D[0];
  int top = 0;
  memset(r, 0, sizeof r);
  for (int i = int(1); i < int(n + 1); i++) {
    while (top > 0 && D[i] > D[st[top - 1]]) {
      r[st[--top]] = i;
    }
    st[top++] = i;
  }
  for (int i = int(1); i < int(n); i++) {
    r[i] -= i + 1;
  }
  top = 0;
  memset(l, 0, sizeof l);
  for (int i = n - 1; i >= 0; i--) {
    while (top > 0 && D[i] >= D[st[top - 1]]) {
      l[st[--top]] = i;
    }
    st[top++] = i;
  }
  for (int i = int(1); i < int(n); i++) {
    l[i] = i - l[i] - 1;
  }
  long long a, b, L, R;
  for (int i = int(0); i < int(q); i++) {
    cin >> a >> b;
    b--;
    long long ans = 0;
    for (int j = int(a); j < int(b + 1); j++) {
      L = min(l[j], j - a);
      R = min(r[j], b - j);
      L++, R++;
      ans += L * R * D[j];
    }
    cout << ans << '\n';
  }
}

### Prompt

Generate 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;
const long long maxN = 1e6 + 5, MOD = 1e9 + 7, INF = 1e9;
const double PI = acos(-1), eps = 1e-7;
int n, q, A[maxN], D[maxN];
long long r[maxN], l[maxN], st[maxN];
int main() {
  cin >> n >> q;
  for (int i = int(1); i < int(n + 1); i++) {
    cin >> A[i];
  }
  A[0] = A[n + 1] = INF;
  for (int i = int(1); i < int(n + 1); i++) {
    D[i - 1] = abs(A[i] - A[i - 1]);
  }
  D[n] = D[0];
  int top = 0;
  memset(r, 0, sizeof r);
  for (int i = int(1); i < int(n + 1); i++) {
    while (top > 0 && D[i] > D[st[top - 1]]) {
      r[st[--top]] = i;
    }
    st[top++] = i;
  }
  for (int i = int(1); i < int(n); i++) {
    r[i] -= i + 1;
  }
  top = 0;
  memset(l, 0, sizeof l);
  for (int i = n - 1; i >= 0; i--) {
    while (top > 0 && D[i] >= D[st[top - 1]]) {
      l[st[--top]] = i;
    }
    st[top++] = i;
  }
  for (int i = int(1); i < int(n); i++) {
    l[i] = i - l[i] - 1;
  }
  long long a, b, L, R;
  for (int i = int(0); i < int(q); i++) {
    cin >> a >> b;
    b--;
    long long ans = 0;
    for (int j = int(a); j < int(b + 1); j++) {
      L = min(l[j], j - a);
      R = min(r[j], b - j);
      L++, R++;
      ans += L * R * D[j];
    }
    cout << ans << '\n';
  }
}

```