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;
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long llread() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long Power(long long x, long long y, long long mod) {
  long long r = 1;
  while (y) {
    if (y & 1) r = r * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return r;
}
int T[110], L[200010 * 60], R[200010 * 60], ans[200010], tot;
inline void add(int &o, int l, int r, int x) {
  if (!o) o = ++tot;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (x <= mid)
    add(L[o], l, mid, x);
  else
    add(R[o], mid + 1, r, x);
}
inline void merge(int &p, int &q, int l, int r, int x, int y) {
  if (!p) return;
  if (!q && x <= l && r <= y) {
    q = p;
    p = 0;
    return;
  }
  if (!q) q = ++tot;
  int mid = (l + r) / 2;
  if (x <= mid) merge(L[p], L[q], l, mid, x, y);
  if (y > mid) merge(R[p], R[q], mid + 1, r, x, y);
  if (!L[p] && !R[p]) p = 0;
  if (!L[q] && !R[q]) q = 0;
}
inline void dfs(int o, int l, int r, int v) {
  if (!o) return;
  if (l == r) {
    ans[l] = v;
    return;
  }
  int mid = (l + r) / 2;
  dfs(L[o], l, mid, v);
  dfs(R[o], mid + 1, r, v);
}
inline void Getans(int n) {
  for (long long i = (1); i <= (100); i++) dfs(T[i], 1, n, i);
  for (long long i = (1); i <= (n); i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int n = read();
  for (long long i = (1); i <= (n); i++) add(T[read()], 1, n, i);
  int q = read();
  while (q--) {
    int l = read(), r = read(), x = read(), y = read();
    if (x == y) continue;
    merge(T[x], T[y], 1, n, l, r);
  }
  Getans(n);
  return 0;
}
