#include<bits/stdc++.h>
using namespace std;
template <typename T,typename E>
struct SegmentTree{
  typedef function<T(T,T)> F;
  typedef function<T(T,E)> G;
  typedef function<E(E,E)> H;
  typedef function<E(E,int)> P;
  int n;
  F f;
  G g;
  H h;
  P p;
  T d1;
  E d0;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(int n_,F f,G g,H h,P p,T d1,E d0){
    this->f=f;
    this->g=g;
    this->h=h;
    this->p=p;
    this->d1=d1;
    this->d0=d0;
    init(n_);
  }
  SegmentTree(int n_,F f,G g,H h,P p,T d1,E d0,vector<T> v){
    this->f=f;
    this->g=g;
    this->h=h;
    this->p=p;
    this->d1=d1;
    this->d0=d0;
    init(n_);
    build(n_,v);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(2*n-1,d1);
    laz.clear();
    laz.resize(2*n-1,d0);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  inline void eval(int len,int k){
    if(k*2+1<n*2-1){
      laz[k*2+1]=h(laz[k*2+1],laz[k]);
      laz[k*2+2]=h(laz[k*2+2],laz[k]);
    }
    dat[k]=g(dat[k],p(laz[k],len));
    laz[k]=d0;
  }
  T update(int a,int b,E x,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return g(dat[k],p(laz[k],r-l));
    if(a<=l&&r<=b){
      laz[k]=h(laz[k],x);
      return g(dat[k],p(laz[k],r-l));
    }
    return dat[k]=f(update(a,b,x,k*2+1,l,(l+r)/2),
		    update(a,b,x,k*2+2,(l+r)/2,r));
  }
  T update(int a,int b,E x){
    return update(a,b,x,0,0,n);
  }
  T query(int a,int b,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return d1;
    if(a<=l&&r<=b) return dat[k];
    T vl=query(a,b,k*2+1,l,(l+r)/2);
    T vr=query(a,b,k*2+2,(l+r)/2,r);
    return f(vl,vr);
  }
  T query(int a,int b){
    return query(a,b,0,0,n);
  }
  void update(int k,T x){
    k+=n-1;
    dat[k]=x;
    while(k){
      k=(k-1)/2;
      dat[k]=f(dat[k*2+1],dat[k*2+2]);
    }
  }
};
signed main(){
  int n;
  scanf("%d",&n);
  typedef pair<int,int> P;
  vector<P> v(n);
  for(int i=0;i<n;i++) scanf("%d",&v[i].first);
  for(int i=0;i<n;i++) scanf("%d",&v[i].second);
  SegmentTree<P,P> rmq(n,
		       [](P a,P b){
			 return P(min(a.first,b.first),
				  min(a.second,b.second));
		       },
		       [](P a,P b){
			 if(b.first<0){
			   if(b.second<0) return a;
			   return P(a.second,a.second);
			 }
			 if(b.second<0)
			   return P(a.first,a.first);
			 if(b.second<b.first)
			   return P(a.second,a.second);
			 return P(a.first,a.first);
		       },
		       [](P a,P b){
			 int x=a.first,y=a.second;
			 if(x<0) x=b.first;
			 if(y<0) y=b.second;
			 return P(x,y);
		       },
		       [](P a,int k){
			 return a;
		       },
		       P(INT_MAX,INT_MAX),P(-1,-1),v);
  int q;
  scanf("%d",&q);
  for(int i=0;i<q;i++){
    int x,y,z;
    scanf("%d %d %d",&x,&y,&z);
    y--;
    if(x==1){
      assert(0<=y&&y<n);
      P c=rmq.query(y,y+1);
      c.first=z;
      rmq.update(y,c);
    }
    if(x==2){
      assert(0<=y&&y<n);
      P c=rmq.query(y,y+1);
      c.second=z;
      rmq.update(y,c);
    }
    if(x==3){
      assert(0<=y&&y<z);
      P c=rmq.query(y,z);
      printf("%d\n",c.first);
    }
    if(x==4){
      assert(0<=y&&y<z);
      P c=rmq.query(y,z);
      printf("%d\n",c.second);
    }
    if(x==5){
      rmq.query(0,rmq.n);
      rmq.update(0,rmq.n,P(-1,i));
    }
    if(x==6){
      rmq.query(0,rmq.n);
      rmq.update(0,rmq.n,P(i,-1));
    }
  }
  return 0;
}
### Prompt

Create a solution in CPP for 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;
template <typename T,typename E>
struct SegmentTree{
  typedef function<T(T,T)> F;
  typedef function<T(T,E)> G;
  typedef function<E(E,E)> H;
  typedef function<E(E,int)> P;
  int n;
  F f;
  G g;
  H h;
  P p;
  T d1;
  E d0;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(int n_,F f,G g,H h,P p,T d1,E d0){
    this->f=f;
    this->g=g;
    this->h=h;
    this->p=p;
    this->d1=d1;
    this->d0=d0;
    init(n_);
  }
  SegmentTree(int n_,F f,G g,H h,P p,T d1,E d0,vector<T> v){
    this->f=f;
    this->g=g;
    this->h=h;
    this->p=p;
    this->d1=d1;
    this->d0=d0;
    init(n_);
    build(n_,v);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(2*n-1,d1);
    laz.clear();
    laz.resize(2*n-1,d0);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  inline void eval(int len,int k){
    if(k*2+1<n*2-1){
      laz[k*2+1]=h(laz[k*2+1],laz[k]);
      laz[k*2+2]=h(laz[k*2+2],laz[k]);
    }
    dat[k]=g(dat[k],p(laz[k],len));
    laz[k]=d0;
  }
  T update(int a,int b,E x,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return g(dat[k],p(laz[k],r-l));
    if(a<=l&&r<=b){
      laz[k]=h(laz[k],x);
      return g(dat[k],p(laz[k],r-l));
    }
    return dat[k]=f(update(a,b,x,k*2+1,l,(l+r)/2),
		    update(a,b,x,k*2+2,(l+r)/2,r));
  }
  T update(int a,int b,E x){
    return update(a,b,x,0,0,n);
  }
  T query(int a,int b,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return d1;
    if(a<=l&&r<=b) return dat[k];
    T vl=query(a,b,k*2+1,l,(l+r)/2);
    T vr=query(a,b,k*2+2,(l+r)/2,r);
    return f(vl,vr);
  }
  T query(int a,int b){
    return query(a,b,0,0,n);
  }
  void update(int k,T x){
    k+=n-1;
    dat[k]=x;
    while(k){
      k=(k-1)/2;
      dat[k]=f(dat[k*2+1],dat[k*2+2]);
    }
  }
};
signed main(){
  int n;
  scanf("%d",&n);
  typedef pair<int,int> P;
  vector<P> v(n);
  for(int i=0;i<n;i++) scanf("%d",&v[i].first);
  for(int i=0;i<n;i++) scanf("%d",&v[i].second);
  SegmentTree<P,P> rmq(n,
		       [](P a,P b){
			 return P(min(a.first,b.first),
				  min(a.second,b.second));
		       },
		       [](P a,P b){
			 if(b.first<0){
			   if(b.second<0) return a;
			   return P(a.second,a.second);
			 }
			 if(b.second<0)
			   return P(a.first,a.first);
			 if(b.second<b.first)
			   return P(a.second,a.second);
			 return P(a.first,a.first);
		       },
		       [](P a,P b){
			 int x=a.first,y=a.second;
			 if(x<0) x=b.first;
			 if(y<0) y=b.second;
			 return P(x,y);
		       },
		       [](P a,int k){
			 return a;
		       },
		       P(INT_MAX,INT_MAX),P(-1,-1),v);
  int q;
  scanf("%d",&q);
  for(int i=0;i<q;i++){
    int x,y,z;
    scanf("%d %d %d",&x,&y,&z);
    y--;
    if(x==1){
      assert(0<=y&&y<n);
      P c=rmq.query(y,y+1);
      c.first=z;
      rmq.update(y,c);
    }
    if(x==2){
      assert(0<=y&&y<n);
      P c=rmq.query(y,y+1);
      c.second=z;
      rmq.update(y,c);
    }
    if(x==3){
      assert(0<=y&&y<z);
      P c=rmq.query(y,z);
      printf("%d\n",c.first);
    }
    if(x==4){
      assert(0<=y&&y<z);
      P c=rmq.query(y,z);
      printf("%d\n",c.second);
    }
    if(x==5){
      rmq.query(0,rmq.n);
      rmq.update(0,rmq.n,P(-1,i));
    }
    if(x==6){
      rmq.query(0,rmq.n);
      rmq.update(0,rmq.n,P(i,-1));
    }
  }
  return 0;
}
```