Problem: Alex decided to try his luck in TV shows. He once went to the quiz named "What's That Word?!". After perfectly answering the questions "How is a pseudonym commonly referred to in the Internet?" ("Um... a nick?"), "After which famous inventor we name the unit of the magnetic field strength?" ("Um... Nikola Tesla?") and "Which rock band performs "How You Remind Me"?" ("Um... Nickelback?"), he decided to apply to a little bit more difficult TV show: "What's in This Multiset?!".

The rules of this TV show are as follows: there are n multisets numbered from 1 to n. Each of them is initially empty. Then, q events happen; each of them is in one of the four possible types:

  * 1 x v — set the x-th multiset to a singleton \\{v\}.
  * 2 x y z — set the x-th multiset to a union of the y-th and the z-th multiset. For example: \{1, 3\}∪\{1, 4, 4\}=\{1, 1, 3, 4, 4\}.
  * 3 x y z — set the x-th multiset to a product of the y-th and the z-th multiset. The product A × B of two multisets A, B is defined as \{ \gcd(a, b)  ∣  a ∈ A,  b ∈ B \}, where \gcd(p, q) is the greatest common divisor of p and q. For example: \{2, 2, 3\} × \{1, 4, 6\}=\{1, 2, 2, 1, 2, 2, 1, 1, 3\}.
  * 4 x v — the participant is asked how many times number v occurs in the x-th multiset. As the quiz turned out to be too hard in the past, participants should now give the answers modulo 2 only. 



Note, that x, y and z described above are not necessarily different. In events of types 2 and 3, the sum or the product is computed first, and then the assignment is performed.

Alex is confused by the complicated rules of the show. Can you help him answer the requests of the 4-th type?

Input

The first line contains two integers n and q (1 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^6) — the number of multisets and the number of events.

Each of the following q lines describes next event in the format given in statement. It's guaranteed that 1 ≤ x,y,z ≤ n and 1 ≤ v ≤ 7000 always holds.

It's guaranteed that there will be at least one event of the 4-th type.

Output

Print a string which consists of digits 0 and 1 only, and has length equal to the number of events of the 4-th type. The i-th digit of the string should be equal to the answer for the i-th query of the 4-th type.

Example

Input


4 13
1 1 1
1 2 4
1 3 6
4 4 4
1 4 4
2 2 1 2
2 3 3 4
4 4 4
3 2 2 3
4 2 1
4 2 2
4 2 3
4 2 4


Output


010101

Note

Here is how the multisets look in the example test after each of the events; i is the number of queries processed so far:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int mxv = 7009;
bitset<mxv> mu, a[100009], mask[71];
int p[7009];
vector<int> g[7009];
void pre() {
  mu.reset();
  mu[1] = 1;
  int cnt = 0;
  for (int i = 2; i <= (7000); ++i) {
    if (!p[i]) {
      for (int j = 2 * i; j <= 7000; j += i) p[j] = i;
      mu[i] = 1;
      cnt++;
    } else {
      if (mu[i / p[i]] == 0 || (i / p[i]) % p[i] == 0)
        mu[i] = 0;
      else
        mu[i] = 1, cnt++;
    }
  }
  for (int i = 1; i <= (7000); ++i) {
    for (int j = i; j <= 7000; j += i) g[j].push_back(i);
  }
  for (int i = 1; i <= (70); ++i) {
    for (int j = i; j <= 7000; j += i) mask[i][j] = mu[j / i];
  }
}
void solve() {}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  pre();
  int n;
  cin >> n;
  int q;
  cin >> q;
  for (int i = 0; i < (q); ++i) {
    int t, x, v;
    cin >> t >> x >> v;
    if (t == 1) {
      a[x].reset();
      for (auto& j : g[v]) a[x][j] = 1;
    } else if (t == 2) {
      int z;
      cin >> z;
      a[x] = a[v] ^ a[z];
    } else if (t == 3) {
      int z;
      cin >> z;
      a[x] = a[v] & a[z];
    } else {
      if (v > 70) {
        bool ans = 0;
        for (int j = 1; j <= (7000); ++j) {
          if (j * v > 7000) break;
          ans ^= mu[j] & a[x][j * v];
        }
        cout << ans;
      } else {
        cout << (mask[v] & a[x]).count() % 2;
      }
    }
  }
  return 0;
}