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;
class SegTree {
 protected:
  struct node {
    int l, r;
    short val, tag[101];
    node() {
      for (register int i = 1; i <= 100; i++) tag[i] = i;
    }
  } tr[200002 << 2];
  void spread(int p) {
    for (register int i = 1; i <= 100; i++) {
      tr[p << 1].tag[i] = tr[p].tag[tr[p << 1].tag[i]];
      tr[p << 1 | 1].tag[i] = tr[p].tag[tr[p << 1 | 1].tag[i]];
    }
    for (register int i = 1; i <= 100; i++) tr[p].tag[i] = i;
  }

 public:
  void build(short a[], int l, int r, int p = 1) {
    if (l == r) {
      tr[p].l = tr[p].r = l;
      tr[p].val = a[l];
      return;
    }
    tr[p].l = l, tr[p].r = r;
    int mid = (l + r) >> 1;
    build(a, l, mid, p << 1);
    build(a, mid + 1, r, p << 1 | 1);
  }
  void change(int l, int r, short x, short y, int p = 1) {
    if (l <= tr[p].l && tr[p].r <= r) {
      for (register int i = 1; i <= 100; i++)
        if (tr[p].tag[i] == x) tr[p].tag[i] = y;
      return;
    }
    spread(p);
    int mid = (tr[p].l + tr[p].r) >> 1;
    if (l <= mid) change(l, r, x, y, p << 1);
    if (mid < r) change(l, r, x, y, p << 1 | 1);
  }
  void ask(int p = 1) {
    if (tr[p].l == tr[p].r) {
      cout << tr[p].tag[tr[p].val] << ' ';
      return;
    }
    spread(p);
    ask(p << 1);
    ask(p << 1 | 1);
  }
} a;
short val[200002];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> val[i];
  a.build(val, 1, n);
  int m;
  cin >> m;
  while (m--) {
    int l, r, x, y;
    cin >> l >> r >> x >> y;
    a.change(l, r, x, y);
  }
  a.ask();
}
