#include <bits/stdc++.h>
#define REP(i, a, n) for(ll i = ((ll) a); i < ((ll) n); i++)
#define INF (1LL << 60)
using namespace std;
typedef long long ll;

ll N, A, B, Q, X, Y, Z;

template<typename T> class SegmentTree {
public:
  ll n;
  T e;
  T (*op)(T, T);
  vector<T> dat;

  SegmentTree(T _e, T(*_op)(T, T), ll _n) {
    e = _e;
    op = _op;
    n = 1;
    while(n < _n) n *= 2;
    REP(i, 0, 2 * n - 1) dat.push_back(e);
  }

  T query(ll a, ll b) { return rquery(a, b, 0, 0, n); };

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

private:
  T rquery(ll a, ll b, ll k, ll l, ll r) {
    if(r <= a || b <= l) return e;
    if(a <= l && r <= b) return dat[k];
    T vl = rquery(a, b, k * 2 + 1, l, (l + r) / 2);
    T vr = rquery(a, b, k * 2 + 2, (l + r) / 2, r);
    return op(vl, vr);
  }
};

ll opmin(ll a, ll b) { return min(a, b); };

int main(void) {
  SegmentTree<ll> sa(INF, opmin, 200000), sb(INF, opmin, 200000);

  cin >> N;
  REP(i, 0, N) {
    cin >> A;
    sa.update(i, A);
    // cout << A << endl;
  }
  REP(i, 0, N) {
    cin >> B;
    sb.update(i, B);
    // cout << B << endl;
  }
  cin >> Q;

  // REP(i, 0, N) cout << sa.dat[i] << " ";
  // cout << endl;
  // REP(i, 0, N) cout << sb.dat[i] << " ";
  // cout << endl;

  vector<ll> u;
  REP(i, 0, N) u.push_back(i);

  // REP(j, 0, N) cout << sa.query(j, j + 1) << " ";
  // cout << endl;
  // REP(j, 0, N) cout << sb.query(j, j + 1) << " ";
  // cout << endl;

  REP(q, 0, Q) {
    cin >> X >> Y >> Z;
    // cout << "  " << X << " " << Y << " " << Z << endl;
    if(X == 1) { // update A[Y] = Z
      sa.update(Y - 1, Z);
      u.push_back(Y - 1);
    }
    if(X == 2) { // update B[Y] = Z
      sb.update(Y - 1, Z);
      u.push_back(Y - 1);
    }
    if(X == 3) { // max(A[Y], ..., A[Z])
      cout << sa.query(Y - 1, Z) << endl;;
    }
    if(X == 4) { // max(B[Y], ..., B[Z])
      cout << sb.query(Y - 1, Z) << endl;
    }
    if(X == 5) { // copy A <- B
      REP(i, 0, u.size()) sa.update(u[i], sb.dat[u[i] + sb.n - 1]);
      u.clear();
    }
    if(X == 6) { // copy B <- A
      REP(i, 0, u.size()) sb.update(u[i], sa.dat[u[i] + sa.n - 1]);
      u.clear();
    }
    // REP(j, 0, N) cout << sa.query(j, j + 1) << " ";
    // cout << endl;
    // REP(j, 0, N) cout << sb.query(j, j + 1) << " ";
    // cout << endl;
  }
}
### 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>
#define REP(i, a, n) for(ll i = ((ll) a); i < ((ll) n); i++)
#define INF (1LL << 60)
using namespace std;
typedef long long ll;

ll N, A, B, Q, X, Y, Z;

template<typename T> class SegmentTree {
public:
  ll n;
  T e;
  T (*op)(T, T);
  vector<T> dat;

  SegmentTree(T _e, T(*_op)(T, T), ll _n) {
    e = _e;
    op = _op;
    n = 1;
    while(n < _n) n *= 2;
    REP(i, 0, 2 * n - 1) dat.push_back(e);
  }

  T query(ll a, ll b) { return rquery(a, b, 0, 0, n); };

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

private:
  T rquery(ll a, ll b, ll k, ll l, ll r) {
    if(r <= a || b <= l) return e;
    if(a <= l && r <= b) return dat[k];
    T vl = rquery(a, b, k * 2 + 1, l, (l + r) / 2);
    T vr = rquery(a, b, k * 2 + 2, (l + r) / 2, r);
    return op(vl, vr);
  }
};

ll opmin(ll a, ll b) { return min(a, b); };

int main(void) {
  SegmentTree<ll> sa(INF, opmin, 200000), sb(INF, opmin, 200000);

  cin >> N;
  REP(i, 0, N) {
    cin >> A;
    sa.update(i, A);
    // cout << A << endl;
  }
  REP(i, 0, N) {
    cin >> B;
    sb.update(i, B);
    // cout << B << endl;
  }
  cin >> Q;

  // REP(i, 0, N) cout << sa.dat[i] << " ";
  // cout << endl;
  // REP(i, 0, N) cout << sb.dat[i] << " ";
  // cout << endl;

  vector<ll> u;
  REP(i, 0, N) u.push_back(i);

  // REP(j, 0, N) cout << sa.query(j, j + 1) << " ";
  // cout << endl;
  // REP(j, 0, N) cout << sb.query(j, j + 1) << " ";
  // cout << endl;

  REP(q, 0, Q) {
    cin >> X >> Y >> Z;
    // cout << "  " << X << " " << Y << " " << Z << endl;
    if(X == 1) { // update A[Y] = Z
      sa.update(Y - 1, Z);
      u.push_back(Y - 1);
    }
    if(X == 2) { // update B[Y] = Z
      sb.update(Y - 1, Z);
      u.push_back(Y - 1);
    }
    if(X == 3) { // max(A[Y], ..., A[Z])
      cout << sa.query(Y - 1, Z) << endl;;
    }
    if(X == 4) { // max(B[Y], ..., B[Z])
      cout << sb.query(Y - 1, Z) << endl;
    }
    if(X == 5) { // copy A <- B
      REP(i, 0, u.size()) sa.update(u[i], sb.dat[u[i] + sb.n - 1]);
      u.clear();
    }
    if(X == 6) { // copy B <- A
      REP(i, 0, u.size()) sb.update(u[i], sa.dat[u[i] + sa.n - 1]);
      u.clear();
    }
    // REP(j, 0, N) cout << sa.query(j, j + 1) << " ";
    // cout << endl;
    // REP(j, 0, N) cout << sb.query(j, j + 1) << " ";
    // cout << endl;
  }
}
```