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 <bits/stdc++.h>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;

const int MOD_TYPE = 1;
const ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
const int INF = (int)1e9;
const ll LINF = (ll)4e18;
const double PI = acos(-1.0);

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define possible(n) cout << ((n) ? "possible" : "impossible") << endl
#define Yay(n) cout << ((n) ? "Yay!" : ":(") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << endl;

vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(50) << setiosflags(ios::fixed);
  int n;
  cin >> n;
  vector<ll> a(n);
  rep(i, n) cin >> a[i];
  int q;
  cin >> q;
  rep(qi, q)
  {
    int b, e;
    cin >> b >> e;
    reverse(a.begin() + b, a.begin() + e);
  }
  rep(i, a.size()) cout << a[i] << (i == a.size() - 1 ? "\n" : " ");
  return 0;
}