#include <iostream>

using namespace std;

int p[10010];
int n,q;
int c,x,y;

int root(int x)
{
    if(p[x]==x)
    {
        return x;
    }
    else
    {
        p[x]=root(p[x]);
        return p[x];
    }
}

void unite(int x,int y)
{
    p[root(x)]=root(y);
}

bool same(int x,int y)
{
    return root(x)==root(y);
}

int main()
{
    cin >> n >> q;
    for(int i=0;i<n;i++)
    {
        p[i]=i;
    }
    for(int i=0;i<q;i++)
    {
        cin >> c >> x >> y;
        if(c)
        {
            cout << same(x,y) << endl;
        }
        else
        {
            unite(x,y);
        }
    }
    return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <iostream>

using namespace std;

int p[10010];
int n,q;
int c,x,y;

int root(int x)
{
    if(p[x]==x)
    {
        return x;
    }
    else
    {
        p[x]=root(p[x]);
        return p[x];
    }
}

void unite(int x,int y)
{
    p[root(x)]=root(y);
}

bool same(int x,int y)
{
    return root(x)==root(y);
}

int main()
{
    cin >> n >> q;
    for(int i=0;i<n;i++)
    {
        p[i]=i;
    }
    for(int i=0;i<q;i++)
    {
        cin >> c >> x >> y;
        if(c)
        {
            cout << same(x,y) << endl;
        }
        else
        {
            unite(x,y);
        }
    }
    return 0;
}

```