#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
int print_value(A title, B val) {
  cout << title << " = " << val << "\n";
  return 1;
}
int break_point() {
  char c;
  while ((c = getchar()) != '\n')
    ;
  return 0;
}
template <typename A>
int logg(A v) {
  cout << v << "\n";
  return 0;
}
template <typename T>
void read_integer(T &r) {
  bool sign = 0;
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c == '-') {
      sign = 1;
      break;
    }
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') break;
    r = r * 10 + (c - '0');
  }
  if (sign) r = -r;
}
void run();
int main() {
  srand(time(NULL));
  do {
    run();
    if (0) {
      0 ? printf("-------------------------------\n") : 0;
      0 ? printf("-------------------------------\n") : 0;
    }
  } while (0);
  return 0;
}
int a[100005];
int D[100005];
int l[100005];
int r[100005];
void run() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (i) D[i] = abs(a[i] - a[i - 1]);
  }
  stack<int> st;
  for (int i = 1; i < n; i++) {
    while (!st.empty() && D[st.top()] < D[i]) st.pop();
    if (st.empty())
      l[i] = 0;
    else
      l[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = n - 1; i >= 1; i--) {
    while (!st.empty() && D[st.top()] <= D[i]) st.pop();
    if (st.empty())
      r[i] = n;
    else
      r[i] = st.top();
    st.push(i);
  }
  int a, b;
  while (q--) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    long long res = 0;
    for (int i = a + 1; i <= b; i++) {
      long long cnt = 1ll * (i - max(l[i], a)) * (min(r[i], b + 1) - i);
      res += cnt * D[i];
    }
    printf("%lld", res);
    putchar('\n');
  }
}

### Prompt

Your challenge is to write 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 <typename A, typename B>
int print_value(A title, B val) {
  cout << title << " = " << val << "\n";
  return 1;
}
int break_point() {
  char c;
  while ((c = getchar()) != '\n')
    ;
  return 0;
}
template <typename A>
int logg(A v) {
  cout << v << "\n";
  return 0;
}
template <typename T>
void read_integer(T &r) {
  bool sign = 0;
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c == '-') {
      sign = 1;
      break;
    }
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') break;
    r = r * 10 + (c - '0');
  }
  if (sign) r = -r;
}
void run();
int main() {
  srand(time(NULL));
  do {
    run();
    if (0) {
      0 ? printf("-------------------------------\n") : 0;
      0 ? printf("-------------------------------\n") : 0;
    }
  } while (0);
  return 0;
}
int a[100005];
int D[100005];
int l[100005];
int r[100005];
void run() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (i) D[i] = abs(a[i] - a[i - 1]);
  }
  stack<int> st;
  for (int i = 1; i < n; i++) {
    while (!st.empty() && D[st.top()] < D[i]) st.pop();
    if (st.empty())
      l[i] = 0;
    else
      l[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = n - 1; i >= 1; i--) {
    while (!st.empty() && D[st.top()] <= D[i]) st.pop();
    if (st.empty())
      r[i] = n;
    else
      r[i] = st.top();
    st.push(i);
  }
  int a, b;
  while (q--) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    long long res = 0;
    for (int i = a + 1; i <= b; i++) {
      long long cnt = 1ll * (i - max(l[i], a)) * (min(r[i], b + 1) - i);
      res += cnt * D[i];
    }
    printf("%lld", res);
    putchar('\n');
  }
}

```