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≤105, 1≤q≤106) — 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
inputCopy
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
outputCopy
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:
题意：
你有一些集合，给你4种操作：
1 x v：让这个集合变成只有一个数字 v
2 x,y,z：把y集合和z集合并起来赋给x
3 x,y,z：求y集合的所有元素与z集合的所有元素的gcd，赋给x
4 x,v：求x中有多少个v，奇数输出1，偶数输出0
题解：
很久很久没有接触莫比乌斯了，连别人的题解都看不懂，搞了好久才有一点头绪。
首先由于第3种操作的存在使得这道题不能直接做，我找不到在时限内可以做出来的方法，这个就得用莫比乌斯反演来做。我们设f(x)=G(x)+G(2x)+G(3x)+…这个是倍数反演，那么反演的方程就是
由于它只需要输出0,1，那么-1和1是等价的，所以莫比乌斯函数就是0,1中的一个值。然后根据这个方程，我们需要把每一个数的所有因子存下来，然后再把每个数的倍数的莫比乌斯函数存下来。
那么第一种操作我们就存v的所有因子，第二种操作就是异或，第三种操作因为我们存的是因子，那么就只需要与操作就好了，第四种，我们需要让集合x与上v的倍数莫比乌斯函数，这样就代表反演方程了。
#include<bits/stdc++.h>
using namespace std;
const int N=7e3+5;
int mu[N],p[N],np[N],cnt;
bitset<N>bs[(int)1e5+5],yz[N],m[N];
int main()
{
mu[1]=1;
for(int i=2;i<N;i++)
{
if(!np[i])
{
p[++cnt]=i;
mu[i]=1;
}
for(int j=1;j<=cnt&&p[j]*i<N;j++)
{
np[p[j]*i]=1;
if(i%p[j]==0)
{
mu[p[j]*i]=0;
break;
}
mu[p[j]*i]^=mu[i];
}
}
for(int i=1;i<N;i++)
{
for(int j=1;j<N;j++)
{
if(i%j==0)
yz[i][j]=1;
if(j%i==0)
m[i][j]=mu[j/i];
}
}
int n,t;
scanf("%d%d",&n,&t);
while(t--)
{
int op,x,y,z;
scanf("%d",&op);
if(op==1)
scanf("%d%d",&x,&y),bs[x]=yz[y];
else if(op==2)
scanf("%d%d%d",&x,&y,&z),bs[x]=bs[y]^bs[z];
else if(op==3)
scanf("%d%d%d",&x,&y,&z),bs[x]=bs[y]&bs[z];
else
scanf("%d%d",&x,&y),printf("%d",(bs[x]&m[y]).count()%2);
}
printf("\n");
return 0;
}