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:
class UnionFind():
  def __init__(self, n):
    self.n = n
    self.parent = [i for i in range(self.n)]
    self.rank = [0] * self.n
  
  def find_root(self, x):
    if self.parent[x] == x:
      return x
    else:
      self.parent[x] = self.find_root(self.parent[x])
      return self.parent[x]
  
  def unite(self, x, y):
    rx = self.find_root(x)
    ry = self.find_root(y)
    if rx == ry:
      return
    if self.rank[rx] < self.rank[ry]:
      self.parent[rx] = ry
    else:
      self.parent[ry] = rx
      if self.rank[rx] == self.rank[ry]:
        self.rank[rx] += 1
  
  def is_same(self, x, y):
    return self.find_root(x) == self.find_root(y)

N, Q = map(int, input().split())
uf = UnionFind(N)
for _ in range(Q):
  p, a, b = map(int, input().split())
  if p == 0:
    uf.unite(a, b)
  else:
    if uf.is_same(a, b):
      print(1)
    else:
      print(0)