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>
#define rep(i,a,n) for(int i=a; i<n; i++)
using namespace std;

const int S = 10010;
int uf[S];

int find(int a) {
    if(a == uf[a]) return a;
    return find(uf[a]);
}

void unite(int a, int b) {
    int fa = find(a);
    int fb = find(b);
    if(fa == fb) return;
    if(fa < fb) uf[fb] = uf[fa];
    else uf[fa] = uf[fb];
}

int main() {
    rep(i,0,S) uf[i] = i;
    int n, q; cin >> n >> q;
    rep(i,0,q) {
        int c, x, y; cin >> c >> x >> y;
        if(c == 0) unite(x, y);
        else if(c == 1) cout << (find(x) == find(y)) << endl;
    }
    return 0;
}