#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF 1e16

struct RangeMinQuery{
  ll dat[(1<<19)-1];
  int size;

  void init(int n_){
    size=1;
    while(size<n_) size*=2;
    for(int i=0;i<2*size-1;i++)dat[i]=INF;
  }

  void update(int k,ll a){
    k+=size-1;
    dat[k]=a;
    while(k>0){
      k=(k-1)/2;
      dat[k]=min(dat[k*2+1],dat[k*2+2]);
    }
  }

  ll subquery(int a,int b,int k,int l,int r){
    if(r<=a||b<=l)return INF;
    if(a<=l&&r<=b)return dat[k];
    else{
      return min(subquery(a,b,k*2+1,l,(l+r)/2),subquery(a,b,k*2+2,(l+r)/2,r));
    }
  }
  
  ll query(int a,int b){
    return subquery(a,b,0,0,size);
  }
};

ll n;
ll a[201010],b[201010];
vector<ll> ch;
RangeMinQuery rmq[2];

int main(){
  cin>>n;
  rep(i,n)cin>>a[i];
  rep(i,n)cin>>b[i];
  rmq[0].init(n); rmq[1].init(n);
  ll q;
  cin>>q;
  bool oc=false;
  rep(i,n)rmq[0].update(i,a[i]);
  rep(i,n)rmq[1].update(i,b[i]);
  while(q--){
    ll x;
    cin>>x;
    if(x==1||x==2){
      x--;
      ll y,z;
      cin>>y>>z;
      y--;
      if(!oc){        
        rmq[x].update(y,z);
      }else{
        rmq[x].update(y,z);
        ch.push_back(y);
      }
    }else if(x==3||x==4){
      ll y,z;
      cin>>y>>z;
      x-=3;
      y--;
      cout<<rmq[x].query(y,z)<<endl;
    }else{
      x-=5;
      ll y,z;
      cin>>y>>z;
      if(!oc){
        oc=true;
        rep(i,n){
          rmq[x].update(i,rmq[1-x].query(i,i+1));
        }
      }else{
        rep(i,ch.size()){
          rmq[x].update(ch[i],rmq[1-x].query(ch[i],ch[i]+1));
        }
        ch.clear();
      }
    }
  }
	return 0;
}
### Prompt

Please formulate a CPP solution 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>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF 1e16

struct RangeMinQuery{
  ll dat[(1<<19)-1];
  int size;

  void init(int n_){
    size=1;
    while(size<n_) size*=2;
    for(int i=0;i<2*size-1;i++)dat[i]=INF;
  }

  void update(int k,ll a){
    k+=size-1;
    dat[k]=a;
    while(k>0){
      k=(k-1)/2;
      dat[k]=min(dat[k*2+1],dat[k*2+2]);
    }
  }

  ll subquery(int a,int b,int k,int l,int r){
    if(r<=a||b<=l)return INF;
    if(a<=l&&r<=b)return dat[k];
    else{
      return min(subquery(a,b,k*2+1,l,(l+r)/2),subquery(a,b,k*2+2,(l+r)/2,r));
    }
  }
  
  ll query(int a,int b){
    return subquery(a,b,0,0,size);
  }
};

ll n;
ll a[201010],b[201010];
vector<ll> ch;
RangeMinQuery rmq[2];

int main(){
  cin>>n;
  rep(i,n)cin>>a[i];
  rep(i,n)cin>>b[i];
  rmq[0].init(n); rmq[1].init(n);
  ll q;
  cin>>q;
  bool oc=false;
  rep(i,n)rmq[0].update(i,a[i]);
  rep(i,n)rmq[1].update(i,b[i]);
  while(q--){
    ll x;
    cin>>x;
    if(x==1||x==2){
      x--;
      ll y,z;
      cin>>y>>z;
      y--;
      if(!oc){        
        rmq[x].update(y,z);
      }else{
        rmq[x].update(y,z);
        ch.push_back(y);
      }
    }else if(x==3||x==4){
      ll y,z;
      cin>>y>>z;
      x-=3;
      y--;
      cout<<rmq[x].query(y,z)<<endl;
    }else{
      x-=5;
      ll y,z;
      cin>>y>>z;
      if(!oc){
        oc=true;
        rep(i,n){
          rmq[x].update(i,rmq[1-x].query(i,i+1));
        }
      }else{
        rep(i,ch.size()){
          rmq[x].update(ch[i],rmq[1-x].query(ch[i],ch[i]+1));
        }
        ch.clear();
      }
    }
  }
	return 0;
}
```