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 <string>
#include <iostream>
#include <stdio.h>
#include <vector>

using namespace std;

void reverse(vector<int>::iterator first, vector<int>::iterator last) {
  for (; first != last && first != --last; ++first) {
    iter_swap(first, last);
  }
}

int main() {
  ios::sync_with_stdio(false);

  int len_array;
  cin >> len_array;
  vector<int> v(len_array);

  for (int i = 0; i < len_array; i++) {
    cin >> v[i];
  }

  int num_query;
  cin >> num_query;

  int b, e;
  for (int i = 0; i < num_query; i++) {
    cin >> b >> e;
    reverse(v.begin() + b, v.begin() + e);
  }

  cout << *v.begin();
  for (auto it = v.begin() + 1, e = v.end(); it != e; ++it) {
    cout << " " << *it;
  }
  cout << endl;
}