#include <bits/stdc++.h>
const int oo = 0x3f3f3f3f;
template <typename T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return a > b ? a = b, true : false;
}
template <typename T>
T read(T &first) {
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (first = 0; isdigit(ch); ch = getchar()) first = 10 * first + ch - '0';
  return first *= f;
}
template <typename T>
void write(T first) {
  if (first < 0) {
    putchar('-');
    first = -first;
  }
  static char s[20];
  int top = 0;
  for (; first; first /= 10) s[++top] = first % 10 + '0';
  if (top == 0)
    putchar('0');
  else
    while (top) putchar(s[top--]);
}
const int MAXN = 1e5 + 5;
int N, Q;
long long A[MAXN];
void input() {
  read(N);
  read(Q);
  for (int i = 1; i <= N; ++i) {
    read(A[i]);
  }
}
void solve() {
  static long long D[MAXN];
  for (int i = 1; i < N; ++i) {
    D[i] = std::abs(A[i] - A[i + 1]);
  }
  while (Q--) {
    int l, r;
    read(l);
    read(r);
    static long long lb[MAXN], ub[MAXN];
    static long long st[MAXN];
    int top = 0;
    long long bak = D[l - 1];
    D[l - 1] = LONG_LONG_MAX;
    st[++top] = l - 1;
    for (int i = l; i < r; ++i) {
      while (D[st[top]] < D[i]) {
        ub[st[top--]] = i;
      }
      lb[i] = st[top] + 1;
      st[++top] = i;
    }
    while (top > 0) {
      ub[st[top--]] = r;
    }
    D[l - 1] = bak;
    long long ans = 0;
    for (int i = l; i < r; ++i) {
      ans += (ub[i] - i) * (i - lb[i] + 1) * D[i];
    }
    write(ans);
    putchar('\n');
  }
}
int main() {
  input();
  solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const int oo = 0x3f3f3f3f;
template <typename T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return a > b ? a = b, true : false;
}
template <typename T>
T read(T &first) {
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (first = 0; isdigit(ch); ch = getchar()) first = 10 * first + ch - '0';
  return first *= f;
}
template <typename T>
void write(T first) {
  if (first < 0) {
    putchar('-');
    first = -first;
  }
  static char s[20];
  int top = 0;
  for (; first; first /= 10) s[++top] = first % 10 + '0';
  if (top == 0)
    putchar('0');
  else
    while (top) putchar(s[top--]);
}
const int MAXN = 1e5 + 5;
int N, Q;
long long A[MAXN];
void input() {
  read(N);
  read(Q);
  for (int i = 1; i <= N; ++i) {
    read(A[i]);
  }
}
void solve() {
  static long long D[MAXN];
  for (int i = 1; i < N; ++i) {
    D[i] = std::abs(A[i] - A[i + 1]);
  }
  while (Q--) {
    int l, r;
    read(l);
    read(r);
    static long long lb[MAXN], ub[MAXN];
    static long long st[MAXN];
    int top = 0;
    long long bak = D[l - 1];
    D[l - 1] = LONG_LONG_MAX;
    st[++top] = l - 1;
    for (int i = l; i < r; ++i) {
      while (D[st[top]] < D[i]) {
        ub[st[top--]] = i;
      }
      lb[i] = st[top] + 1;
      st[++top] = i;
    }
    while (top > 0) {
      ub[st[top--]] = r;
    }
    D[l - 1] = bak;
    long long ans = 0;
    for (int i = l; i < r; ++i) {
      ans += (ub[i] - i) * (i - lb[i] + 1) * D[i];
    }
    write(ans);
    putchar('\n');
  }
}
int main() {
  input();
  solve();
  return 0;
}

```