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:
// 失敗するからこそ そこから立ち向かって行く強さがあってそんな強さが本当の強さだと私は思うから
// ぜったいあきらめない
#include<bits/stdc++.h>
using namespace std;

#if defined(DAIJOBU)
#include "/home/ichigo/debug.hpp"
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define deb(x...)
#endif
#define FIO {ios::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);}



int main(){
	FIO
	
	int n;
	cin >> n;
	int a[n];
	for(int i = 0; i < n; i++)cin >> a[i];
	int q;
	cin >> q;
	while(q--){
		int b, e;
		cin >> b >> e;
		reverse(a + b, a + e);
	}
	cout << a[0];
	for(int i = 1; i < n; i++)cout << ' ' << a[i];
	cout << '\n';
	return 0;
}