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 int N = 1 << 18;
int a[N];
int p[N * 2 + 5][100];
int l, r, x, y;
void push(int i) {
  if (i < N) {
    for (int j = 0; j < 100; j++) {
      p[i * 2][j] = p[i][p[i * 2][j]];
      p[i * 2 + 1][j] = p[i][p[i * 2 + 1][j]];
    }
    for (int j = 0; j < 100; j++) {
      p[i][j] = j;
    }
  }
}
void update(int i, int L, int R) {
  if (R <= l || L >= r) return;
  if (l <= L && R <= r) {
    for (int j = 0; j < 100; j++) {
      if (p[i][j] == x) {
        p[i][j] = y;
      }
    }
    return;
  }
  push(i);
  int m = (L + R) >> 1;
  update(i * 2, L, m);
  update(i * 2 + 1, m, R);
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 1; i < N * 2; i++) {
    for (int j = 0; j < 100; j++) {
      p[i][j] = j;
    }
  }
  int m;
  cin >> m;
  while (m--) {
    cin >> l >> r >> x >> y;
    l--, x--, y--;
    update(1, 0, N);
  }
  for (int i = 1; i < N * 2; i++) push(i);
  for (int i = 0; i < n; i++) {
    cout << p[i + N][a[i]] + 1 << ' ';
  }
  cout << endl;
  return 0;
}
