#include <iostream>
#include <stdio.h>
#define MAXN 10010
using namespace std;
int pre[MAXN];
int find(int x)
{
	return (pre[x] == x) ? x : pre[x] = find(pre[x]);
}
int main(void)
{
	int n, q;

	while (scanf("%d%d", &n, &q) != EOF)
	{
		bool op;
		int x, y;

		for (int i = 0; i < n; i++)pre[i] = i;
		for (int i = 0; i < q; i++)
		{
			scanf("%d%d%d", &op, &x, &y);
			x = find(x), y = find(y);
			if (op)printf("%d\n", x == y);
			else pre[x] = y;
		}
	}

	return 0;
}

### Prompt

Generate 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>
#include <stdio.h>
#define MAXN 10010
using namespace std;
int pre[MAXN];
int find(int x)
{
	return (pre[x] == x) ? x : pre[x] = find(pre[x]);
}
int main(void)
{
	int n, q;

	while (scanf("%d%d", &n, &q) != EOF)
	{
		bool op;
		int x, y;

		for (int i = 0; i < n; i++)pre[i] = i;
		for (int i = 0; i < q; i++)
		{
			scanf("%d%d%d", &op, &x, &y);
			x = find(x), y = find(y);
			if (op)printf("%d\n", x == y);
			else pre[x] = y;
		}
	}

	return 0;
}

```