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:
from typing import List

class UnionFind:
	def __init__(self, V: int) -> None:
		self.V: int = V
		self.pi: List[int] = [i for i in range(V)]
		self.rank: int = [0 for _ in range(V)]

	def findSet(self, x: int) -> int:
		if x != self.pi[x]:
			self.pi[x] = self.findSet(self.pi[x])
		return self.pi[x]

	def same(self, x: int, y: int) -> bool:
		return self.findSet(x) == self.findSet(y)

	def link(self, x: int, y: int) -> None:
		if self.rank[x] > self.rank[y]:
			self.pi[y] = x
		else:
			self.pi[x] = y
			if self.rank[x] == self.rank[y]:
				self.rank[y] += 1

	def unite(self, x: int, y: int) -> None:
		self.link(self.findSet(x), self.findSet(y))


if __name__ == "__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:
			if UF.same(x, y):
				print(1)
			else:
				print(0)