You are given an array a consisting of n integers. You have to process q queries to this array; each query is given as four numbers l, r, x and y, denoting that for every i such that l ≤ i ≤ r and ai = x you have to set ai equal to y.

Print the array after all queries are processed.

Input

The first line contains one integer n (1 ≤ n ≤ 200000) — the size of array a.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100) — the elements of array a.

The third line contains one integer q (1 ≤ q ≤ 200000) — the number of queries you have to process.

Then q lines follow. i-th line contains four integers l, r, x and y denoting i-th query (1 ≤ l ≤ r ≤ n, 1 ≤ x, y ≤ 100).

Output

Print n integers — elements of array a after all changes are made.

Example

Input

5
1 2 3 4 5
3
3 5 3 5
1 5 5 1
1 5 1 5


Output

5 2 5 4 5 
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int N = 2e5 + 5, sz = 500;
int n, arr[N], sq;
list<int> b[sz][105];
void updatel(int li, int l, int r, int x, int y) {
  for (auto it = b[li][x].begin(); it != b[li][x].end();) {
    auto tmp = it;
    tmp++;
    if (*it >= l && *it <= r) b[li][y].splice(b[li][y].end(), b[li][x], it);
    it = tmp;
  }
}
void update(int l, int r, int x, int y) {
  int li = l / sq, ri = r / sq;
  updatel(li, l, r, x, y);
  if (li == ri) return;
  for (int i = li + 1; i < ri; i++) {
    b[i][y].splice(b[i][y].end(), b[i][x]);
  }
  updatel(ri, l, r, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  sq = ceil(sqrt(n));
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    b[i / sq][arr[i]].push_back(i);
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r, x, y;
    cin >> l >> r >> x >> y;
    if (x == y) continue;
    l--, r--;
    update(l, r, x, y);
  }
  for (int i = 1; i <= 100; i++) {
    for (int j = 0; j * sq < n; j++) {
      if (b[j][i].size()) {
        for (auto it : b[j][i]) arr[it] = i;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << arr[i] << ' ';
  return 0;
}
