#include <bits/stdc++.h>
using namespace std;
template <class X, class Y>
void minimize(X &x, const Y &y) {
  if (x > y) x = y;
}
template <class X, class Y>
void maximize(X &x, const Y &y) {
  if (x < y) x = y;
}
int a[100100], b[100100];
int ___left[100100], ___right[100100];
int n, q;
void init(void) {
  scanf("%d%d", &n, &q);
  for (int i = (1), _b = (n); i <= _b; i = i + 1) scanf("%d", &a[i]);
  for (int i = (1), _b = (n - 1); i <= _b; i = i + 1)
    b[i] = max(a[i + 1] - a[i], a[i] - a[i + 1]);
}
void prepare(void) {
  stack<int> st;
  for (int i = (1), _b = (n - 1); i <= _b; i = i + 1) {
    while (!st.empty() && b[st.top()] <= b[i]) st.pop();
    ___left[i] = st.empty() ? 0 : st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = (n - 1), _a = (1); i >= _a; i = i - 1) {
    while (!st.empty() && b[st.top()] < b[i]) st.pop();
    ___right[i] = st.empty() ? n : st.top();
    while (!st.empty() && b[st.top()] <= b[i]) st.pop();
    st.push(i);
  }
}
long long answer(int l, int r) {
  long long res = 0;
  for (int i = (l), _b = (r - 1); i <= _b; i = i + 1)
    res += 1LL * b[i] * (i - max(___left[i] + 1, l) + 1) *
           (min(___right[i], r) - i);
  return (res);
}
void process(void) {
  for (int love = 0, _n = (q); love < _n; love = love + 1) {
    int l, r;
    scanf("%d%d", &l, &r);
    cout << answer(l, r) << "\n";
  }
}
int main(void) {
  init();
  prepare();
  process();
  return 0;
}

### 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;
template <class X, class Y>
void minimize(X &x, const Y &y) {
  if (x > y) x = y;
}
template <class X, class Y>
void maximize(X &x, const Y &y) {
  if (x < y) x = y;
}
int a[100100], b[100100];
int ___left[100100], ___right[100100];
int n, q;
void init(void) {
  scanf("%d%d", &n, &q);
  for (int i = (1), _b = (n); i <= _b; i = i + 1) scanf("%d", &a[i]);
  for (int i = (1), _b = (n - 1); i <= _b; i = i + 1)
    b[i] = max(a[i + 1] - a[i], a[i] - a[i + 1]);
}
void prepare(void) {
  stack<int> st;
  for (int i = (1), _b = (n - 1); i <= _b; i = i + 1) {
    while (!st.empty() && b[st.top()] <= b[i]) st.pop();
    ___left[i] = st.empty() ? 0 : st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = (n - 1), _a = (1); i >= _a; i = i - 1) {
    while (!st.empty() && b[st.top()] < b[i]) st.pop();
    ___right[i] = st.empty() ? n : st.top();
    while (!st.empty() && b[st.top()] <= b[i]) st.pop();
    st.push(i);
  }
}
long long answer(int l, int r) {
  long long res = 0;
  for (int i = (l), _b = (r - 1); i <= _b; i = i + 1)
    res += 1LL * b[i] * (i - max(___left[i] + 1, l) + 1) *
           (min(___right[i], r) - i);
  return (res);
}
void process(void) {
  for (int love = 0, _n = (q); love < _n; love = love + 1) {
    int l, r;
    scanf("%d%d", &l, &r);
    cout << answer(l, r) << "\n";
  }
}
int main(void) {
  init();
  prepare();
  process();
  return 0;
}

```