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 = 200005;
int n, m, sx, sy, now, ans[N], sz, rt[105];
struct tree {
  int l, r;
} t[N * 20];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void ins(int &d, int l, int r, int x) {
  if (!d) d = ++sz;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (x <= mid)
    ins(t[d].l, l, mid, x);
  else
    ins(t[d].r, mid + 1, r, x);
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  t[x].l = merge(t[x].l, t[y].l);
  t[x].r = merge(t[x].r, t[y].r);
  return x;
}
void modify(int &d, int &p, int l, int r) {
  if (!d) return;
  if (sx <= l && r <= sy) {
    p = merge(d, p);
    d = 0;
    return;
  }
  if (!p) p = ++sz;
  int mid = (l + r) / 2;
  if (sx <= mid) modify(t[d].l, t[p].l, l, mid);
  if (sy > mid) modify(t[d].r, t[p].r, mid + 1, r);
}
void solve(int x, int l, int r) {
  if (!x) return;
  if (l == r) {
    ans[l] = now;
    return;
  }
  int mid = (l + r) / 2;
  solve(t[x].l, l, mid);
  solve(t[x].r, mid + 1, r);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    ins(rt[x], 1, n, i);
  }
  m = read();
  while (m--) {
    sx = read();
    sy = read();
    int x = read(), y = read();
    if (x != y) modify(rt[x], rt[y], 1, n);
  }
  for (int i = 1; i <= 100; i++) now = i, solve(rt[i], 1, n);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
