

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

typedef long long ll;
int n, q;

int par[10000];

int root(int a) {
    if(a == par[a])
        return a;
    return par[a] = root(par[a]);
}

bool same(int a, int b) { return root(a) == root(b); }

void unite(int a, int b) {
    a = root(a);
    b = root(b);
    par[a] = b;
}

int main() {
    cin >> n >> q;
    for(int i = 0; i < n; i++)
        par[i] = i;
    for(int i = 0; i < q; i++) {
        int com, x, y;
        cin >> com >> x >> y;
        if(com == 0)
            unite(x, y);
        else
            cout << (same(x, y) ? 1 : 0) << endl;
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp


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

typedef long long ll;
int n, q;

int par[10000];

int root(int a) {
    if(a == par[a])
        return a;
    return par[a] = root(par[a]);
}

bool same(int a, int b) { return root(a) == root(b); }

void unite(int a, int b) {
    a = root(a);
    b = root(b);
    par[a] = b;
}

int main() {
    cin >> n >> q;
    for(int i = 0; i < n; i++)
        par[i] = i;
    for(int i = 0; i < q; i++) {
        int com, x, y;
        cin >> com >> x >> y;
        if(com == 0)
            unite(x, y);
        else
            cout << (same(x, y) ? 1 : 0) << endl;
    }
}

```