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 $n$ ($1 \le n \le 2 \cdot 10^{5}$) — the numer of citizens.
The next line contains $n$ integers $a_1$, $a_2$, …, $a_n$ ($0 \le a_{i} \le 10^{9}$) — the initial balances of citizens.
The next line contains a single integer $q$ ($1 \le q \le 2 \cdot 10^{5}$) — 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 ($1 \le p \le n$, $0 \le x \le 10^{9}$), or 2 x ($0 \le x \le 10^{9}$). 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 $\rightarrow$ 3 3 3 4 $\rightarrow$ 3 2 3 4 $\rightarrow$ 3 2 3 4
In the second example the balances change as follows: 3 50 2 1 10 $\rightarrow$ 3 0 2 1 10 $\rightarrow$ 8 8 8 8 10 $\rightarrow$ 8 8 20 8 10
Solution:
#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); int n; cin >> n; vector<int> a(n); for (int i = 0; i < n; i++) { cin >> a[i]; } int tt; cin >> tt; vector<int> op(tt); vector<int> who(tt); vector<int> val(tt); for (int i = 0; i < tt; i++) { cin >> op[i]; if (op[i] == 1) { cin >> who[i] >> val[i]; --who[i]; } else { cin >> val[i]; } } vector<int> res(n, -1); int mx = -1; for (int i = tt - 1; i >= 0; i--) { if (op[i] == 1) { if (res[who[i]] == -1) { res[who[i]] = max(val[i], mx); } } else { mx = max(mx, val[i]); } } for (int i = 0; i < n; i++) { if (res[i] == -1) { res[i] = max(a[i], mx); } } for (int i = 0; i < n; i++) { if (i > 0) { cout << " "; } cout << res[i]; } cout << '\n'; return 0; }