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 <algorithm>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <ostream>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
typedef long long int ll;
#define rep(i, a, n) for (int i = (a); i < (n); i++)
#define repr(i, n, a) for (int i = (n)-1; i >= (a); i--)
#define each(it, v) for (auto(it) = (v).begin(); (it) != (v).end(); (it)++)
#define all(v) (v).begin(), (v).end()
#define vi vector<int>
#define vvi vector<vi>
#define vd vector<double>
#define vvd vector<vd>
#define co(x) cout << (x)
#define con(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define out(x) rep(i, 0, x.size()) cout << x[i] << ((i != x.size() - 1) ? " " : "\n")
const double PI = 3.14159265;
const int MAXN = 1 << 15;
const ll LINF = 1LL << 60;
const int INF = 1 << 30;
#define Pr(a) (#a) << "=" << (a) << " "
#define pr(a) cout << Pr(a) << '\n'
///////////////////////////////////////////////////////
int main() {
  int n, q;
  cin >> n;
  vi A(n);
  rep(i, 0, n) cin >> A[i];
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    reverse(A.begin() + a, A.begin() + b);
  }
  out(A);
  return 0;
}