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 <string>
#include <algorithm>
#include <set>
#include <cmath>
#include <vector>
#include <numeric>
#include <unordered_map>
#include <unordered_set>
#include <queue>

typedef long long ll;
const ll LL_MAX (1LL<<60);

#define rep(i,s,e) for(ll i=(s); i<(e); i++) 

using namespace std;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n;
    cin >>n;
    vector<ll> a(n);
    rep(i,0,n) cin >> a[i];
    ll q;
    cin >> q;
    vector<ll> b(q), e(q);
    rep(i,0,q) cin >> b[i] >> e[i];

    rep(i,0,q)
    {
        reverse(&a[b[i]], &a[e[i]]);
    }

    rep(i,0,n)
    {
        if (i!=n-1)
            cout << a[i] << ' '; 
        else
            cout << a[i]; 
    }
    cout << endl;

    return 0;
}