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

Solution (in PYTHON3):

Solution:
import queue


class UnionFind:

    def __init__(self, N):
        self.sizes = [1 for _ in range(N)]
        self.par = [i for i in range(N)]

    def find(self, x):
        if x == self.par[x]:
            return x

        self.par[x] = self.find(self.par[x])

        return self.par[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.sizes[x] < self.sizes[y]:
            x, y = y, x

        self.par[y] = x
        self.sizes[x] += self.sizes[y]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def size(self, x):
        return self.sizes[self.find(x)]


def main():
    N, Q = map(int, input().split())
    uf = UnionFind(N)

    for _ in range(Q):
        com, x, y = map(int, input().split())
        if com == 0:
            uf.unite(x, y)
        else:
            print(int(uf.same(x, y)))

    return


if __name__ == '__main__':
    main()