#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n_=1<<19;

class segtree{
public:
	int dp[1<<20];
	segtree(){
		memset(dp,0,sizeof(dp));
	}
	void update(int k,int v){
		k+=n_-1;
		dp[k]=v;
		while(k>0){
			k=(k-1)/2;
			dp[k]=min(dp[k*2+1],dp[k*2+2]);
		}
	}

	int query(int a,int b,int k=0,int l=0,int r=n_){
		if(b<=l || r<=a)return 2e9;
		if(a<=l && r<=b)return dp[k];
		int mid=(l+r)/2;
		int vl=query(a,b,k*2+1,l,mid);
		int vr=query(a,b,k*2+2,mid,r);
		return min(vl,vr);
	}
};

segtree seg[2];

int n,q;
int a[200005],b[200005];
set<int> s;

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
		seg[0].update(i,a[i]);
	}
	for(int i=0;i<n;i++){
		scanf("%d",&b[i]);
		seg[1].update(i,b[i]);
	}
	scanf("%d",&q);
	bool flag=false;
	for(int i=0;i<q;i++){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		y--;
		if(x==1){
			s.insert(y);
			seg[0].update(y,z);
		}
		if(x==2){
			s.insert(y);
			seg[1].update(y,z);
		}
		if(x==3){
			printf("%d\n",seg[0].query(y,z));
		}
		if(x==4){
			printf("%d\n",seg[1].query(y,z));
		}
		if(x==5 || x==6){
			int side=x-5;
			if(!flag){
				for(int i=0;i<n;i++){
					seg[side].update(i,seg[1-side].query(i,i+1));
				}
				flag=true;
			}else{
				for(set<int>::iterator it=s.begin();it!=s.end();it++){
					seg[side].update(*it,seg[1-side].query(*it,(*it)+1));
				}
			}
			s.clear();
		}
	}
	return 0;
}
### Prompt

Please create a solution in CPP to the following problem:

Problem

Given two sequences of length $ N $, $ A $ and $ B $. First, the $ i $ item in the sequence $ A $ is $ a_i $, and the $ i $ item in the sequence $ B $ is $ b_i $.

Since a total of $ Q $ of statements of the following format are given, create a program that processes in the given order.

Each statement is represented by three integers $ x, y, z $.

* Set the value of the $ y $ item in the sequence $ A $ to $ z $. (When $ x = 1 $)
* Set the value of the $ y $ item in the sequence $ B $ to $ z $. (When $ x = 2 $)
* Find and report the smallest value in the $ z $ item from the $ y $ item in the sequence $ A $. (When $ x = 3 $)
* Find and report the smallest value in the $ z $ item from the $ y $ item in the sequence $ B $. (When $ x = 4 $)
* Change the sequence $ A $ to be exactly the same as the sequence $ B $. (When $ x = 5 $)
* Change the sequence $ B $ to be exactly the same as the sequence $ A $. (When $ x = 6 $)

Constraints

The input satisfies the following conditions.

* $ 2 \ le N \ le 2 \ times 10 ^ 5 $
* $ 2 \ le Q \ le 2 \ times 10 ^ 5 $
* $ 1 \ le a_i \ le 10 ^ 9 $
* $ 1 \ le b_i \ le 10 ^ 9 $
* $ 1 \ le x_i \ le 6 $
* $ 1 \ le y_i \ le N $ (when $ 1 \ le x_i \ le 4 $)
* $ y_i = -1 $ (when $ x_i = 5, 6 $)
* $ 1 \ le z_i \ le 10 ^ 9 $ (when $ x_i = 1, 2 $)
* $ y_i \ le z_i \ le N $ (when $ x_i = 3, 4 $)
* $ z_i = -1 $ (when $ x_i = 5, 6 $)
* All inputs are integers

Input

The input is given in the following format.

$ N $
$ a_ {1} $ $ a_ {2} $ ... $ a_ {N} $
$ b_ {1} $ $ b_ {2} $ ... $ b_ {N} $
$ Q $
$ x_1 $ $ y_1 $ $ z_1 $
$ x_2 $ $ y_2 $ $ z_2 $
...
$ x_Q $ $ y_Q $ $ z_Q $

Output

Every time a statement of $ x = 3 $ or $ x = 4 $ is given by input, the found value is output on one line.

Example

Input

5
1 3 5 7 9
6 2 3 2 6
10
1 3 4
3 4 5
4 2 3
5 -1 -1
2 3 8
3 2 5
4 3 3
1 1 1
6 -1 -1
3 1 5


Output

7
2
2
8
1

### Response

```cpp
#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n_=1<<19;

class segtree{
public:
	int dp[1<<20];
	segtree(){
		memset(dp,0,sizeof(dp));
	}
	void update(int k,int v){
		k+=n_-1;
		dp[k]=v;
		while(k>0){
			k=(k-1)/2;
			dp[k]=min(dp[k*2+1],dp[k*2+2]);
		}
	}

	int query(int a,int b,int k=0,int l=0,int r=n_){
		if(b<=l || r<=a)return 2e9;
		if(a<=l && r<=b)return dp[k];
		int mid=(l+r)/2;
		int vl=query(a,b,k*2+1,l,mid);
		int vr=query(a,b,k*2+2,mid,r);
		return min(vl,vr);
	}
};

segtree seg[2];

int n,q;
int a[200005],b[200005];
set<int> s;

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
		seg[0].update(i,a[i]);
	}
	for(int i=0;i<n;i++){
		scanf("%d",&b[i]);
		seg[1].update(i,b[i]);
	}
	scanf("%d",&q);
	bool flag=false;
	for(int i=0;i<q;i++){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		y--;
		if(x==1){
			s.insert(y);
			seg[0].update(y,z);
		}
		if(x==2){
			s.insert(y);
			seg[1].update(y,z);
		}
		if(x==3){
			printf("%d\n",seg[0].query(y,z));
		}
		if(x==4){
			printf("%d\n",seg[1].query(y,z));
		}
		if(x==5 || x==6){
			int side=x-5;
			if(!flag){
				for(int i=0;i<n;i++){
					seg[side].update(i,seg[1-side].query(i,i+1));
				}
				flag=true;
			}else{
				for(set<int>::iterator it=s.begin();it!=s.end();it++){
					seg[side].update(*it,seg[1-side].query(*it,(*it)+1));
				}
			}
			s.clear();
		}
	}
	return 0;
}
```