Problem: Write a program which reads a sequence of integers $A = \\{a_0, a_1, ..., a_{n-1}\\}$ and reverse specified elements by a list of the following operation:

* reverse($b, e$): reverse the order of $a_b, a_{b+1}, ..., a_{e-1}$

Constraints

* $1 \leq n \leq 1,000$
* $-1,000,000,000 \leq a_i \leq 1,000,000,000$
* $1 \leq q \leq 1,000$
* $0 \leq b < e \leq n$

Input

The input is given in the following format.


$n$
$a_0 \; a_1 \; ...,\; a_{n-1}$
$q$
$b_1 \; e_1$
$b_2 \; e_2$
:
$b_{q} \; b_{q}$


In the first line, $n$ (the number of elements in $A$) is given. In the second line, $a_i$ (each element in $A$) are given. In the third line, the number of queries $q$ is given and each query is given by two integers $b_i \; e_i$ in the following $q$ lines.

Output

Print all elements of $A$ in a line after performing the given operations. Put a single space character between adjacency elements and a newline at the end of the last element.

Example

Input

8
1 2 3 4 5 6 7 8
2
1 6
3 8


Output

1 6 5 8 7 2 3 4

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
  vector<int> a;
  int n,num;

  cin >> n;

  for (int i = 0; i < n; i++) {
    cin >> num;
    a.push_back(num);
    //cout << num << endl;
  }

  int q,b,e;
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> b;
    cin >> e;

    auto itr_b = a.begin();
    auto itr_e = a.begin();

    for (int i=0;i<b;i++){
      itr_b++;
    }
    for (int i=0;i<e;i++){
      itr_e++;
    }
    reverse(itr_b, itr_e++);
  }
  auto itr = a.begin();
  for (itr = a.begin(); itr!=a.end(); ++itr) {
    if (itr!=--a.end()){
      printf("%d ",*itr);
    }else{
      printf("%d\n",*itr);
    }
  }
return 0;
}