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;
#define F first
#define S second
typedef long long lli;
const int MX = 10004 , infi =1000000 , mod = 1000000007;
typedef pair<int,int> pii;

int n , m , fat[MX] , x , y , z;

int fin(int x)
{
    if(x!=fat[x])
        return fat[x] = fin(fat[x]);
    return x;
}

void unio(int a , int b)
{
    int aa = fin(a);
    int bb = fin(b);
    if(aa==bb)return;
    fat[aa]=bb;
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    for(int i=0;i<MX;i++)fat[i]=i;
    cin>>n>>m;
    while(m-- && cin>>x>>y>>z)
    {
        if(x)cout<<(fin(y)==fin(z))<<"\n";
        else unio(y,z);
    }
    return 0;
}