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 #今後も使う値、基本的には全て、self.n に代入する
    self.parents = [-1]*n # -1 で初期化
  def unite(self,x,y):
    x = self.find(x)
    y = self.find(y)
    if x == y: return
    if self.parents[x] > self.parents[y]:
      x,y = y,x
    
    self.parents[x] += self.parents[y]
    self.parents[y] = x
  def find(self,x):#rootを見つける
    if self.parents[x] < 0:
      return x
    else:
      #縮約
      self.parents[x] = self.find(self.parents[x])#再帰！親のさらにnodeを辿った値に更新する。
      return self.parents[x]
      #縮約なしver
      # return self.find(self.parents[x])
  def same(self,a,b):
    return self.find(a) == self.find(b)
if __name__ == "__main__":
  n,q = map(int,input().split())
  uf = UnionFind(n)
  for i in range(q):
    com,x,y = map(str,input().split())
    x = int(x)
    y = int(y)
    if com == "0": 
      uf.unite(x,y)
      #print(uf.parents,i)
    if com == "1": 
      if uf.same(x,y): print(1)
      else: print(0)
  #print(uf.parents)