There is a country with n citizens. The i-th of them initially has a_{i} money. The government strictly controls the wealth of its citizens. Whenever a citizen makes a purchase or earns some money, they must send a receipt to the social services mentioning the amount of money they currently have.

Sometimes the government makes payouts to the poor: all citizens who have strictly less money than x are paid accordingly so that after the payout they have exactly x money. In this case the citizens don't send a receipt.

You know the initial wealth of every citizen and the log of all events: receipts and payouts. Restore the amount of money each citizen has after all events.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the numer of citizens.

The next line contains n integers a_1, a_2, ..., a_n (0 ≤ a_{i} ≤ 10^{9}) — the initial balances of citizens.

The next line contains a single integer q (1 ≤ q ≤ 2 ⋅ 10^{5}) — the number of events.

Each of the next q lines contains a single event. The events are given in chronological order.

Each event is described as either 1 p x (1 ≤ p ≤ n, 0 ≤ x ≤ 10^{9}), or 2 x (0 ≤ x ≤ 10^{9}). In the first case we have a receipt that the balance of the p-th person becomes equal to x. In the second case we have a payoff with parameter x.

Output

Print n integers — the balances of all citizens after all events.

Examples

Input


4
1 2 3 4
3
2 3
1 2 2
2 1


Output


3 2 3 4 


Input


5
3 50 2 1 10
3
1 2 0
2 8
1 3 20


Output


8 8 20 8 10 

Note

In the first example the balances change as follows: 1 2 3 4 → 3 3 3 4 → 3 2 3 4 → 3 2 3 4

In the second example the balances change as follows: 3 50 2 1 10 → 3 0 2 1 10 → 8 8 8 8 10 → 8 8 20 8 10
#include <bits/stdc++.h>
using namespace std;
const int HAUTEUR = 18;
const int TAILLE = (1 << (HAUTEUR + 1));
const int iDebFeuilles = (1 << HAUTEUR);
const int VIDE = 1e9 + 1;
int arbre[TAILLE];
bool prop[TAILLE];
void descente(int pos, int chemin, int prof, int val) {
  if (pos >= iDebFeuilles) {
    arbre[pos] = val;
  } else {
    if (prop[pos]) {
      prop[pos] = false;
      prop[2 * pos] = prop[2 * pos + 1] = true;
      arbre[2 * pos] = max(arbre[2 * pos], arbre[pos]);
      arbre[2 * pos + 1] = max(arbre[2 * pos + 1], arbre[pos]);
    }
    arbre[pos] = min(arbre[pos], val);
    if (chemin & (1 << prof))
      descente(2 * pos + 1, chemin, prof - 1, val);
    else
      descente(2 * pos, chemin, prof - 1, val);
  }
}
int main() {
  for (int i = 0; i < TAILLE; i++) arbre[i] = VIDE;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arbre[iDebFeuilles + i]);
  for (int i = iDebFeuilles - 1; i > 0; i--)
    arbre[i] = min(arbre[2 * i], arbre[2 * i + 1]);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int p, x;
      scanf("%d%d", &p, &x);
      p--;
      descente(1, p, HAUTEUR - 1, x);
    } else {
      int x;
      scanf("%d", &x);
      arbre[1] = max(arbre[1], x);
      prop[1] = true;
    }
  }
  for (int pos = 1; pos < iDebFeuilles; pos++) {
    if (!prop[pos]) continue;
    prop[2 * pos] = prop[2 * pos + 1] = true;
    arbre[2 * pos] = max(arbre[2 * pos], arbre[pos]);
    arbre[2 * pos + 1] = max(arbre[2 * pos + 1], arbre[pos]);
  }
  for (int i = 0; i < n; i++) printf("%d ", arbre[iDebFeuilles + i]);
}
