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 sys


class UnionFindTree:
    def __init__(self, n_node):
        self.parent = [i for i in range(n_node)]
        self.rank = [0] * n_node

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

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

        if x != y:
            if self.rank[x] < self.rank[y]:
                self.parent[x] = y
            else:
                self.parent[y] = x
                if self.rank[x] == self.rank[y]:
                    self.rank[x] += 1

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


if __name__ == "__main__":
    n_node, n_command = map(int, input().split())
    uf_tree = UnionFindTree(n_node)

    for command, x, y in (map(int, i.split()) for i in sys.stdin):
        if command == 0:
            uf_tree.unite(x, y)
        else:
            print(int(uf_tree.have_same_root(x, y)))