Problem:
Disjoint Set

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.

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.

Constraints

1 ≤ n ≤ 10000

1 ≤ q ≤ 100000

x ≠ y

Sample 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

Sample Output

0
0
1
1
1
0
1
1

Solution in C (judge verdict: Accepted)

#include<stdio.h>

void same(int*, int*, int*);
void unite(int*, int*, int*, int*);

int main(void) {
    int i, j;
    int n, q, com, x, y;
    scanf("%d%d", &n, &q);
    int map[n];
    
    for(i=0; i<n; i++) map[i] = i;
    
    for(i=0; i<q; i++) {
        scanf("%d%d%d", &com, &x, &y);
        
        if(com == 1) {
            same(map, &x, &y);
        } else {
            unite(&n, map, &x, &y);
        }
    }
    return 0;
}

void same(int *map, int *x, int*y) {
    if(map[*x] == map[*y]) printf("1\n");
    else printf("0\n");
}

void unite(int *n, int *map, int *x, int *y) {
    int i;
    int target = map[*y];
    for(i=0; i<*n; i++) {
        if(map[i]==target) map[i] = map[*x];
    }
}