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;
struct query {
  int t, x, y;
  query(int tt, int xx, int yy) {
    t = tt;
    x = xx;
    y = yy;
  }
};
vector<int> tree[800005];
vector<query> v[200005];
int arr[200005];
void build(int node, int st, int en) {
  for (int i = 0; i <= 100; i++) tree[node].push_back(i);
  if (st != en) {
    int mid = (st + en) / 2;
    build(2 * node, st, mid);
    build(2 * node + 1, mid + 1, en);
  }
}
void update(int node, int st, int en, query q) {
  if (st == en)
    tree[node][q.x] = q.y;
  else {
    int mid = (st + en) / 2;
    if (st <= q.t && q.t <= mid)
      update(2 * node, st, mid, q);
    else
      update(2 * node + 1, mid + 1, en, q);
    for (int i = 0; i <= 100; i++)
      tree[node][i] = tree[2 * node + 1][tree[2 * node][i]];
  }
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l, r, x, y;
    scanf("%d%d%d%d", &l, &r, &x, &y);
    v[l].push_back(query(i, x, y));
    v[r + 1].push_back(query(i, x, x));
  }
  build(1, 0, q - 1);
  for (int i = 1; i <= n; i++) {
    for (auto x : v[i]) update(1, 0, q - 1, x);
    printf("%d ", tree[1][arr[i]]);
  }
}
