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 UnionFindTree:    
    def __init__(self,n):
        self.a = [i for i in range(n)]
        self.rank = [1 for i in range(n)]

    def root(self,x):
        if self.a[x] != x:
            return self.root(self.a[x])
        return x

    def same(self,x,y):
        if self.root(x) == self.root(y):
            return 1
        else:
            return 0

    def union(self,x,y):
        rx = self.root(x)
        ry = self.root(y)
        if rx != ry:
            if self.rank[rx] < self.rank[ry]:
                self.a[rx] = ry
            else:
                self.a[ry] = rx
                if self.rank[rx] == self.rank[ry]:
                    self.rank[rx] += 1
            
                

n,q = map(int,input().split())
unionFindTree = UnionFindTree(n)

for i in range(q):
    com,x,y = map(int,input().split())
    if com == 0:
        unionFindTree.union(x,y)
    elif com == 1:
        print(unionFindTree.same(x,y))