#include <cstdio>
#include <algorithm>

const int S = 2e5 * 30, inf = 2e9;

int ls[S], rs[S], mx[S], sz;

int build(int l, int r) {
  int o = sz++;
  ls[o] = rs[o] = 0;
  if (l + 1 == r) {
    scanf("%d", &mx[o]);
    return o;
  }
  int m = (l + r) >> 1;
  ls[o] = build(l, m);
  rs[o] = build(m, r);
  mx[o] = std::min(mx[ls[o]], mx[rs[o]]);
  return o;
}

int ins(int o, int l, int r, int x, int v) {
  int p = sz++;
  ls[p] = ls[o];
  rs[p] = rs[o];
  if (l + 1 == r) {
    mx[p] = v;
    return p;
  }
  int m = (l + r) >> 1;
  if (x < m) ls[p] = ins(ls[o], l, m, x, v);
  else rs[p] = ins(rs[o], m, r, x, v);
  mx[p] = std::min(mx[ls[p]], mx[rs[p]]);
  return p;
}

int ask(int o, int l, int r, int L, int R) {
  if (L <= l && R >= r) return mx[o];
  int ret = inf, m = (l + r) >> 1;
  if (L < m) ret = std::min(ret, ask(ls[o], l, m, L, R));
  if (R > m) ret = std::min(ret, ask(rs[o], m, r, L, R));
  return ret;
}

int main() {
  ls[0] = rs[0] = 0;
  mx[0] = inf;
  sz = 1;
  int n;
  scanf("%d", &n);
  int ra = build(0, n);
  int rb = build(0, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) ra = ins(ra, 0, n, y - 1, z);
    if (x == 2) rb = ins(rb, 0, n, y - 1, z);
    if (x == 3) printf("%d\n", ask(ra, 0, n, y - 1, z));
    if (x == 4) printf("%d\n", ask(rb, 0, n, y - 1, z));
    if (x == 5) ra = rb;
    if (x == 6) rb = ra;
  }
  return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <cstdio>
#include <algorithm>

const int S = 2e5 * 30, inf = 2e9;

int ls[S], rs[S], mx[S], sz;

int build(int l, int r) {
  int o = sz++;
  ls[o] = rs[o] = 0;
  if (l + 1 == r) {
    scanf("%d", &mx[o]);
    return o;
  }
  int m = (l + r) >> 1;
  ls[o] = build(l, m);
  rs[o] = build(m, r);
  mx[o] = std::min(mx[ls[o]], mx[rs[o]]);
  return o;
}

int ins(int o, int l, int r, int x, int v) {
  int p = sz++;
  ls[p] = ls[o];
  rs[p] = rs[o];
  if (l + 1 == r) {
    mx[p] = v;
    return p;
  }
  int m = (l + r) >> 1;
  if (x < m) ls[p] = ins(ls[o], l, m, x, v);
  else rs[p] = ins(rs[o], m, r, x, v);
  mx[p] = std::min(mx[ls[p]], mx[rs[p]]);
  return p;
}

int ask(int o, int l, int r, int L, int R) {
  if (L <= l && R >= r) return mx[o];
  int ret = inf, m = (l + r) >> 1;
  if (L < m) ret = std::min(ret, ask(ls[o], l, m, L, R));
  if (R > m) ret = std::min(ret, ask(rs[o], m, r, L, R));
  return ret;
}

int main() {
  ls[0] = rs[0] = 0;
  mx[0] = inf;
  sz = 1;
  int n;
  scanf("%d", &n);
  int ra = build(0, n);
  int rb = build(0, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) ra = ins(ra, 0, n, y - 1, z);
    if (x == 2) rb = ins(rb, 0, n, y - 1, z);
    if (x == 3) printf("%d\n", ask(ra, 0, n, y - 1, z));
    if (x == 4) printf("%d\n", ask(rb, 0, n, y - 1, z));
    if (x == 5) ra = rb;
    if (x == 6) rb = ra;
  }
  return 0;
}
```