#include <bits/stdc++.h>

using namespace std;


template< typename T >
struct SegmentTree
{
  const T INF;

  vector< T > seg;
  int sz;

  SegmentTree(int n) : INF(numeric_limits< T >::max())
  {
    sz = 1;
    while(sz < n) sz <<= 1;
    seg.assign(2 * sz - 1, INF);
  }

  T merge(T a, T b)
  {
    return (min(a, b));
  }

  void set(int k, int x)
  {
    seg[k + sz - 1] = x;
  }

  void build()
  {
    for(int k = sz - 2; k >= 0; k--) {
      seg[k] = merge(seg[2 * k + 1], seg[2 * k + 2]);
    }
  }

  T rmq(int a, int b, int k, int l, int r)
  {
    if(a >= r || b <= l) return (INF);
    if(a <= l && r <= b) return (seg[k]);
    return (merge(rmq(a, b, 2 * k + 1, l, (l + r) >> 1),
                  rmq(a, b, 2 * k + 2, (l + r) >> 1, r)));
  }

  T rmq(int a, int b)
  {
    return (rmq(a, b, 0, 0, sz));
  }

  void update(int k, T x)
  {
    k += sz - 1;
    seg[k] = x;
    while(k > 0) {
      k = (k - 1) >> 1;
      seg[k] = merge(seg[2 * k + 1], seg[2 * k + 2]);
    }
  }
};

int main()
{
  int N, A[200000], B[200000];

  scanf("%d", &N);
  for(int i = 0; i < N; i++) scanf("%d", &A[i]);
  for(int i = 0; i < N; i++) scanf("%d", &B[i]);
  SegmentTree< int > seg1(N), seg2(N);
  for(int i = 0; i < N; i++) seg1.set(i, A[i]);
  for(int i = 0; i < N; i++) seg2.set(i, B[i]);
  seg1.build();
  seg2.build();

  int Q;
  scanf("%d", &Q);
  bool first = false;
  vector< pair< int, int > > aa, bb, cc, dd;
  while(Q--) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    --b;
    if(a == 1) cc.emplace_back(b, seg1.rmq(b, b + 1)), seg1.update(b, c), aa.emplace_back(b, c);
    if(a == 2) dd.emplace_back(b, seg2.rmq(b, b + 1)), seg2.update(b, c), bb.emplace_back(b, c);
    if(a == 3) printf("%d\n", seg1.rmq(b, c));
    if(a == 4) printf("%d\n", seg2.rmq(b, c));
    if(a == 5) {
      if(!first) {
        for(int i = 0; i < N; i++) seg1.update(i, seg2.rmq(i, i + 1));
      } else {
        reverse(begin(cc), end(cc));
        for(auto &p : cc) seg1.update(p.first, p.second);
        for(auto &p : bb) seg1.update(p.first, p.second);
      }
      first = true;
      aa.clear();
      bb.clear();
      cc.clear();
      dd.clear();
    }
    if(a == 6) {
      if(!first) {
        for(int i = 0; i < N; i++) seg2.update(i, seg1.rmq(i, i + 1));
      } else {
        reverse(begin(dd), end(dd));
        for(auto &p : dd) seg2.update(p.first, p.second);
        for(auto &p : aa) seg2.update(p.first, p.second);
      }
      first = true;
      aa.clear();
      bb.clear();
      cc.clear();
      dd.clear();
    }
  }
}
### 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 <bits/stdc++.h>

using namespace std;


template< typename T >
struct SegmentTree
{
  const T INF;

  vector< T > seg;
  int sz;

  SegmentTree(int n) : INF(numeric_limits< T >::max())
  {
    sz = 1;
    while(sz < n) sz <<= 1;
    seg.assign(2 * sz - 1, INF);
  }

  T merge(T a, T b)
  {
    return (min(a, b));
  }

  void set(int k, int x)
  {
    seg[k + sz - 1] = x;
  }

  void build()
  {
    for(int k = sz - 2; k >= 0; k--) {
      seg[k] = merge(seg[2 * k + 1], seg[2 * k + 2]);
    }
  }

  T rmq(int a, int b, int k, int l, int r)
  {
    if(a >= r || b <= l) return (INF);
    if(a <= l && r <= b) return (seg[k]);
    return (merge(rmq(a, b, 2 * k + 1, l, (l + r) >> 1),
                  rmq(a, b, 2 * k + 2, (l + r) >> 1, r)));
  }

  T rmq(int a, int b)
  {
    return (rmq(a, b, 0, 0, sz));
  }

  void update(int k, T x)
  {
    k += sz - 1;
    seg[k] = x;
    while(k > 0) {
      k = (k - 1) >> 1;
      seg[k] = merge(seg[2 * k + 1], seg[2 * k + 2]);
    }
  }
};

int main()
{
  int N, A[200000], B[200000];

  scanf("%d", &N);
  for(int i = 0; i < N; i++) scanf("%d", &A[i]);
  for(int i = 0; i < N; i++) scanf("%d", &B[i]);
  SegmentTree< int > seg1(N), seg2(N);
  for(int i = 0; i < N; i++) seg1.set(i, A[i]);
  for(int i = 0; i < N; i++) seg2.set(i, B[i]);
  seg1.build();
  seg2.build();

  int Q;
  scanf("%d", &Q);
  bool first = false;
  vector< pair< int, int > > aa, bb, cc, dd;
  while(Q--) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    --b;
    if(a == 1) cc.emplace_back(b, seg1.rmq(b, b + 1)), seg1.update(b, c), aa.emplace_back(b, c);
    if(a == 2) dd.emplace_back(b, seg2.rmq(b, b + 1)), seg2.update(b, c), bb.emplace_back(b, c);
    if(a == 3) printf("%d\n", seg1.rmq(b, c));
    if(a == 4) printf("%d\n", seg2.rmq(b, c));
    if(a == 5) {
      if(!first) {
        for(int i = 0; i < N; i++) seg1.update(i, seg2.rmq(i, i + 1));
      } else {
        reverse(begin(cc), end(cc));
        for(auto &p : cc) seg1.update(p.first, p.second);
        for(auto &p : bb) seg1.update(p.first, p.second);
      }
      first = true;
      aa.clear();
      bb.clear();
      cc.clear();
      dd.clear();
    }
    if(a == 6) {
      if(!first) {
        for(int i = 0; i < N; i++) seg2.update(i, seg1.rmq(i, i + 1));
      } else {
        reverse(begin(dd), end(dd));
        for(auto &p : dd) seg2.update(p.first, p.second);
        for(auto &p : aa) seg2.update(p.first, p.second);
      }
      first = true;
      aa.clear();
      bb.clear();
      cc.clear();
      dd.clear();
    }
  }
}
```