#include <bits/stdc++.h>
using namespace std;
const int MAX_N = (int)(1e5) + 10;
const int MAX_NODE_NUM = MAX_N;
int power[30];
int st[MAX_NODE_NUM * 2][32];
int ln[MAX_NODE_NUM * 2];
void init_RMQ(int f[], int n) {
  int i, j;
  for (power[0] = 1, i = 1; i < 31; i++) {
    power[i] = 2 * power[i - 1];
  }
  for (i = 0; i < n; i++) {
    st[i][0] = i;
  }
  ln[0] = -1;
  for (int i = 1; i <= n; i++) {
    ln[i] = ln[i >> 1] + 1;
  }
  for (j = 1; j <= ln[n]; j++) {
    for (i = 0; i < n; i++) {
      if (i + power[j - 1] - 1 >= n) {
        break;
      }
      if (f[st[i][j - 1]] < f[st[i + power[j - 1]][j - 1]]) {
        st[i][j] = st[i + power[j - 1]][j - 1];
      } else {
        st[i][j] = st[i][j - 1];
      }
    }
  }
}
int query(int f[], int x, int y) {
  if (x > y) {
    swap(x, y);
  }
  int k = ln[y - x + 1];
  if (f[st[x][k]] < f[st[y - power[k] + 1][k]]) return st[y - power[k] + 1][k];
  return st[x][k];
}
int n, q;
int f[MAX_N];
int diff[MAX_N];
long long work(int l, int r) {
  if (l > r) return 0;
  if (l == r) {
    return diff[l];
  }
  long long div = query(diff, l, r);
  long long ret = (div - l + 1) * (r - div + 1) * diff[div];
  ret += work(l, div - 1);
  ret += work(div + 1, r);
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &f[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    diff[i] = abs(f[i] - f[i + 1]);
  }
  init_RMQ(diff, n - 1);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b -= 2;
    printf("%I64d\n", work(a, b));
  }
  return 0;
}

### Prompt

Your task is to create 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 int MAX_N = (int)(1e5) + 10;
const int MAX_NODE_NUM = MAX_N;
int power[30];
int st[MAX_NODE_NUM * 2][32];
int ln[MAX_NODE_NUM * 2];
void init_RMQ(int f[], int n) {
  int i, j;
  for (power[0] = 1, i = 1; i < 31; i++) {
    power[i] = 2 * power[i - 1];
  }
  for (i = 0; i < n; i++) {
    st[i][0] = i;
  }
  ln[0] = -1;
  for (int i = 1; i <= n; i++) {
    ln[i] = ln[i >> 1] + 1;
  }
  for (j = 1; j <= ln[n]; j++) {
    for (i = 0; i < n; i++) {
      if (i + power[j - 1] - 1 >= n) {
        break;
      }
      if (f[st[i][j - 1]] < f[st[i + power[j - 1]][j - 1]]) {
        st[i][j] = st[i + power[j - 1]][j - 1];
      } else {
        st[i][j] = st[i][j - 1];
      }
    }
  }
}
int query(int f[], int x, int y) {
  if (x > y) {
    swap(x, y);
  }
  int k = ln[y - x + 1];
  if (f[st[x][k]] < f[st[y - power[k] + 1][k]]) return st[y - power[k] + 1][k];
  return st[x][k];
}
int n, q;
int f[MAX_N];
int diff[MAX_N];
long long work(int l, int r) {
  if (l > r) return 0;
  if (l == r) {
    return diff[l];
  }
  long long div = query(diff, l, r);
  long long ret = (div - l + 1) * (r - div + 1) * diff[div];
  ret += work(l, div - 1);
  ret += work(div + 1, r);
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &f[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    diff[i] = abs(f[i] - f[i + 1]);
  }
  init_RMQ(diff, n - 1);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b -= 2;
    printf("%I64d\n", work(a, b));
  }
  return 0;
}

```