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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long mxn = 200005;
const int mod = 1000000007;
unsigned char arr[mxn + 1];
void solve_case() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%hhu", &arr[i]);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    unsigned char x, y;
    scanf("%d %d %hhu %hhu", &l, &r, &x, &y);
    for (int i = l; i <= r; i++) {
      if (arr[i] == x) arr[i] = y;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%hhu ", arr[i]);
  }
  printf("\n");
  return;
}
int main() {
  solve_case();
  return 0;
}
