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:
#UnionFind
class UnionFind(object):
  def __init__(self, n=1):
    self.par = [i for i in range(n)]
    self.rank = [0 for _ in range(n)]

  def find(self, x):
    #x が属するグループを探索
    if self.par[x] == x:
      return x
    else:
      self.par[x] = self.find(self.par[x])
      return self.par[x]

  def union(self, x, y):
    #x と y のグループを結合
    x = self.find(x)
    y = self.find(y)
    if x != y:
      if self.rank[x] < self.rank[y]:
        x, y = y, x
      if self.rank[x] == self.rank[y]:
        self.rank[x] += 1
      self.par[y] = x

  def is_same(self, x, y):
    #x と y が同じグループか否か
    return self.find(x) == self.find(y)
#uf=UnionFind(N) uf.union(A,B)のように使う

N, Q = map(int, input().split())

uf = UnionFind(N)
for _ in range(Q):
    P, A, B = map(int, input().split())
    # 連結クエリ
    if P == 0:
        uf.union(A, B)
    # 判定クエリ
    else:
        if uf.is_same(A, B):
            print(1)
        else:
            print(0)