#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
template <typename T>
bool chkmin(T &a, T b) {
  return b < a ? a = b, true : false;
}
template <typename T>
bool chkmax(T &a, T b) {
  return b > a ? a = b, true : false;
}
template <class T>
void read(T &first) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  first = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') first = first * 10 + c - '0';
  first *= flag;
}
template <class T>
void write(T first) {
  if (first < 0) putchar('-'), first = -first;
  static char s[20];
  int top = 0;
  while (first) s[++top] = first % 10 + '0', first /= 10;
  if (top == 0) putchar('0');
  while (top) putchar(s[top--]);
}
const int MAXN = 1e5 + 10;
int a[MAXN], L[MAXN], R[MAXN];
int N, M;
int st[MAXN];
int main() {
  read(N), read(M);
  for (int i = 1; i <= N; ++i) read(a[i]);
  for (int i = N; i >= 1; --i) a[i] = abs(a[i] - a[i - 1]);
  int top = 0;
  for (int i = 2; i <= N; ++i) {
    while (top && a[st[top]] <= a[i]) --top;
    if (top) L[i] = st[top] + 1;
    st[++top] = i;
  }
  top = 0;
  for (int i = N; i >= 2; --i) {
    while (top && a[st[top]] < a[i]) --top;
    if (top)
      R[i] = st[top] - 1;
    else
      R[i] = N;
    st[++top] = i;
  }
  while (M--) {
    int l, r;
    long long ans = 0;
    read(l), read(r);
    l++;
    for (int i = l; i <= r; ++i) {
      int left = max(l, L[i]), right = min(r, R[i]);
      ans += (long long)(right - i + 1) * (i - left + 1) * a[i];
    }
    write(ans), putchar('\n');
  }
  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 oo = 0x3f3f3f3f;
template <typename T>
bool chkmin(T &a, T b) {
  return b < a ? a = b, true : false;
}
template <typename T>
bool chkmax(T &a, T b) {
  return b > a ? a = b, true : false;
}
template <class T>
void read(T &first) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  first = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') first = first * 10 + c - '0';
  first *= flag;
}
template <class T>
void write(T first) {
  if (first < 0) putchar('-'), first = -first;
  static char s[20];
  int top = 0;
  while (first) s[++top] = first % 10 + '0', first /= 10;
  if (top == 0) putchar('0');
  while (top) putchar(s[top--]);
}
const int MAXN = 1e5 + 10;
int a[MAXN], L[MAXN], R[MAXN];
int N, M;
int st[MAXN];
int main() {
  read(N), read(M);
  for (int i = 1; i <= N; ++i) read(a[i]);
  for (int i = N; i >= 1; --i) a[i] = abs(a[i] - a[i - 1]);
  int top = 0;
  for (int i = 2; i <= N; ++i) {
    while (top && a[st[top]] <= a[i]) --top;
    if (top) L[i] = st[top] + 1;
    st[++top] = i;
  }
  top = 0;
  for (int i = N; i >= 2; --i) {
    while (top && a[st[top]] < a[i]) --top;
    if (top)
      R[i] = st[top] - 1;
    else
      R[i] = N;
    st[++top] = i;
  }
  while (M--) {
    int l, r;
    long long ans = 0;
    read(l), read(r);
    l++;
    for (int i = l; i <= r; ++i) {
      int left = max(l, L[i]), right = min(r, R[i]);
      ans += (long long)(right - i + 1) * (i - left + 1) * a[i];
    }
    write(ans), putchar('\n');
  }
  return 0;
}

```