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:
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
#include <array>
#include <queue>
#include <deque>
#include <set>
#include <list>
#include <map>
#include <stack>
#include <utility>
#include <algorithm>
#include <numeric>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <cstring>
#include <climits>
#include <bitset>
#include <random>
#include <functional>
#include <sstream>
#include <iomanip>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); i++)
#define FOR(i, m, n) for(int i=(m);i<(n);i++)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define SORT(x) sort((x).begin(),(x).end())
#define REVE(x) reverse((x).begin(),(x).end())
#define mp make_pair
#define pb push_back
#define F first
#define S second


typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<vector<int>> VVI;
typedef pair<int, int> PII;
typedef long long LL;

int main() {
	int n, q;
	cin >> n;
	VI a(n); rep(i, n)cin >> a[i];
	cin >> q;
	rep(i, q) {
		int b, e;
		cin >> b >> e;
		reverse(a.begin() + b, a.begin() + e);
	}
	rep(i, sz(a)) {
		cout << a[i];
		printf(i == sz(a) - 1 ? "\n" : " ");
	}
}