Problem: Write a program which manipulates a disjoint set S = {S1, S2, . . . , Sk}.

First of all, the program should read an integer n, then make a disjoint set where each element consists of 0, 1, ... n−1 respectively.

Next, the program should read an integer q and manipulate the set for q queries. There are two kinds of queries for different operations:

* unite(x, y): unites sets that contain x and y, say Sx and Sy, into a new set.
* same(x, y): determine whether x and y are in the same set.

Constraints

* 1 ≤ n ≤ 10000
* 1 ≤ q ≤ 100000
* x ≠ y

Input


n q
com1 x1 y1
com2 x2 y2
...
comq xq yq


In the first line, n and q are given. Then, q queries are given where com represents the type of queries. '0' denotes unite and '1' denotes same operation.

Output

For each same operation, print 1 if x and y are in the same set, otherwise 0, in a line.

Example

Input

5 12
0 1 4
0 2 3
1 1 2
1 3 4
1 1 4
1 3 2
0 1 3
1 2 4
1 3 0
0 0 4
1 0 2
1 3 0


Output

0
0
1
1
1
0
1
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

struct UnionFind 
{
vector<int> par;
UnionFind(int N) : par(N)
{
for(int i=0;i<N;i++)par[i]=i;
}
int root(int x)
{
if(par[x]==x)return x;
return par[x]=root(par[x]);
}
void unite(int x,int y)
{
int rx=root(x);int ry=root(y);
if(rx==ry)return;
par[rx]=ry;
}
bool same(int x,int y)
{
int rx=root(x);int ry=root(y);
return rx==ry;
}
};

int main()
{
int n,q;scanf("%d %d",&n,&q);
UnionFind tree(n);
for(int i=0,com,x,y;i<q;i++)
{
scanf("%d %d %d",&com,&x,&y);
if(com==0)tree.unite(x,y);
else if(tree.same(x,y))printf("1\n");
else printf("0\n");
}
}