#include <stdio.h>
#define N 10010
int n;
int father[N];
int findfather(int x){
    return x == father[x] ? x : findfather(father[x]);
}
void init(){
    for(int i = 0; i < n; i++) father[i] = i;
}
int main(){
    int m;
    scanf("%d %d", &n, &m);
    init();
    for(int i = 0; i < m; i++){
        int a, b, c;
        scanf("%d %d %d", &a, &b, &c);
        if(a == 0){
            father[findfather(c)] = findfather(b);
        }
        else{
            printf("%s\n", findfather(b) == findfather(c) ? "1" : "0");
        }
    }
    return 0;
}


### Prompt

Develop a solution in Cpp to the problem described below:

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 <stdio.h>
#define N 10010
int n;
int father[N];
int findfather(int x){
    return x == father[x] ? x : findfather(father[x]);
}
void init(){
    for(int i = 0; i < n; i++) father[i] = i;
}
int main(){
    int m;
    scanf("%d %d", &n, &m);
    init();
    for(int i = 0; i < m; i++){
        int a, b, c;
        scanf("%d %d %d", &a, &b, &c);
        if(a == 0){
            father[findfather(c)] = findfather(b);
        }
        else{
            printf("%s\n", findfather(b) == findfather(c) ? "1" : "0");
        }
    }
    return 0;
}


```