D. Welfare State

There is a country with 𝑛n citizens. The 𝑖i-th of them initially has 𝑎𝑖ai 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 𝑥x are paid accordingly so that after the payout they have exactly 𝑥x 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 𝑛n (1≤𝑛≤2⋅1051≤n≤2⋅105) — the numer of citizens.

The next line contains 𝑛n integers 𝑎1a1, 𝑎2a2, ..., 𝑎𝑛an (0≤𝑎𝑖≤1090≤ai≤109) — the initial balances of citizens.

The next line contains a single integer 𝑞q (1≤𝑞≤2⋅1051≤q≤2⋅105) — the number of events.

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

Each event is described as either 1 p x (1≤𝑝≤𝑛1≤p≤n, 0≤𝑥≤1090≤x≤109), or 2 x (0≤𝑥≤1090≤x≤109). In the first case we have a receipt that the balance of the 𝑝p-th person becomes equal to 𝑥x. In the second case we have a payoff with parameter 𝑥x.

Output

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

Examples

input

Copy

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

output

Copy

3 2 3 4 

input

Copy

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

output

Copy

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

 

哭了哭了~~

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200010;
int a[maxn],last[maxn],b[maxn];//a为原数组,last[i]记录修改i节点的最后一个位置,b[i]代表从i时间点往后最大的补充x值
int main()
{
    int n,q;
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    cin>>q;
    for(int i=1;i<=q;i++){
        int op,x,y;
        cin>>op;
        if(op==1){
            cin>>x>>y;
            a[x]=y;//修改原数组值
            last[x]=i;//记录修改x下标的最后一个位置i
        }
        else{
            cin>>b[i];//输入i时间点的补充值b[i]
        }
    }
    for(int i=q-1;i>=0;i--)b[i]=max(b[i],b[i+1]);
    for(int i=1;i<=n;i++)cout<<max(a[i],b[last[i]])<<" ";
    cout<<endl;
    return 0;
}