#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<unordered_map>
#include<array>
#include<map>
#include<iomanip>
//<< setprecision(5)
using namespace std;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
#define dme cout<<-1<<endl;return 0
const int mod=1000000007;
const int big=1e9+10;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-7;
template <class T,class U>void mineq(T& a,U b){if(a>b){a=b;}}
template <class T,class U>void maxeq(T& a,U b){if(a<b){a=b;}}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){return a/gcd(a,b)*b;}
class seg{public:
	int val;
	int seghi,segmg;
	seg *left,*right;
	seg(int hi,int mg){
		left=nullptr;right=nullptr;val=0;
		seghi=hi;segmg=mg;
		if(hi+1==mg){return;}
		left=new seg(hi,(hi+mg)/2);
		right=new seg((hi+mg)/2,mg);
	}
	int ask(int hi,int mg){
		if(hi<=seghi&&segmg<=mg){return val;}
		if(mg<=seghi||segmg<=hi){return big;}
		return min(left->ask(hi,mg),right->ask(hi,mg));
	}
	void kou(int bas,int num){
		if(seghi==bas&&seghi+1==segmg){val=num;return;}
		if(bas<(seghi+segmg)/2){left->kou(bas,num);}
		else{right->kou(bas,num);}
		val=min(left->val,right->val);
	}
};
int main(void){
	int i,j,n;cin>>n;
	seg A(0,n);
	seg B(0,n);
	vector<int>re;//??´??°?????????????¨??????????????????´??°????????????
	for(i=0;i<n;i++){int a;cin>>a;A.kou(i,a);}
	for(i=0;i<n;i++){int a;cin>>a;B.kou(i,a);}
	for(i=0;i<n;i++){re.pub(i);}
	int q;cin>>q;
	while(q--){
		int x,y,z;cin>>x>>y>>z;
		if(x==1){A.kou(y-1,z);re.pub(y-1);}
		if(x==2){B.kou(y-1,z);re.pub(y-1);}
		if(x==3){cout<<A.ask(y-1,z)<<endl;}
		if(x==4){cout<<B.ask(y-1,z)<<endl;}
		if(x==5){
			for(i=0;i<re.size();i++){
				A.kou(re[i],B.ask(re[i],re[i]+1));
			}
			re.clear();
		}
		if(x==6){
			for(i=0;i<re.size();i++){
				B.kou(re[i],A.ask(re[i],re[i]+1));
			}
			re.clear();
		}
	}
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<unordered_map>
#include<array>
#include<map>
#include<iomanip>
//<< setprecision(5)
using namespace std;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
#define dme cout<<-1<<endl;return 0
const int mod=1000000007;
const int big=1e9+10;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-7;
template <class T,class U>void mineq(T& a,U b){if(a>b){a=b;}}
template <class T,class U>void maxeq(T& a,U b){if(a<b){a=b;}}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){return a/gcd(a,b)*b;}
class seg{public:
	int val;
	int seghi,segmg;
	seg *left,*right;
	seg(int hi,int mg){
		left=nullptr;right=nullptr;val=0;
		seghi=hi;segmg=mg;
		if(hi+1==mg){return;}
		left=new seg(hi,(hi+mg)/2);
		right=new seg((hi+mg)/2,mg);
	}
	int ask(int hi,int mg){
		if(hi<=seghi&&segmg<=mg){return val;}
		if(mg<=seghi||segmg<=hi){return big;}
		return min(left->ask(hi,mg),right->ask(hi,mg));
	}
	void kou(int bas,int num){
		if(seghi==bas&&seghi+1==segmg){val=num;return;}
		if(bas<(seghi+segmg)/2){left->kou(bas,num);}
		else{right->kou(bas,num);}
		val=min(left->val,right->val);
	}
};
int main(void){
	int i,j,n;cin>>n;
	seg A(0,n);
	seg B(0,n);
	vector<int>re;//??´??°?????????????¨??????????????????´??°????????????
	for(i=0;i<n;i++){int a;cin>>a;A.kou(i,a);}
	for(i=0;i<n;i++){int a;cin>>a;B.kou(i,a);}
	for(i=0;i<n;i++){re.pub(i);}
	int q;cin>>q;
	while(q--){
		int x,y,z;cin>>x>>y>>z;
		if(x==1){A.kou(y-1,z);re.pub(y-1);}
		if(x==2){B.kou(y-1,z);re.pub(y-1);}
		if(x==3){cout<<A.ask(y-1,z)<<endl;}
		if(x==4){cout<<B.ask(y-1,z)<<endl;}
		if(x==5){
			for(i=0;i<re.size();i++){
				A.kou(re[i],B.ask(re[i],re[i]+1));
			}
			re.clear();
		}
		if(x==6){
			for(i=0;i<re.size();i++){
				B.kou(re[i],A.ask(re[i],re[i]+1));
			}
			re.clear();
		}
	}
	return 0;
}
```