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._tree = [i for i in range(n)]
        self._rank = [1] * n
    def root(self, a):
        if self._tree[a] == a:return a
        self._tree[a] = self.root(self._tree[a])
        return self._tree[a]
    def is_same_set(self, a, b):
        return self.root(a)==self.root(b)
    def unite(self, a, b):
        ra = self.root(a)
        rb = self.root(b)
        if ra == rb: return
        if self._rank[ra] < self._rank[rb]:
            self._tree[ra] = rb
        else:
            self._tree[rb] = ra
            if self._rank[ra] == self._rank[rb]:
                self._rank[ra] += 1

n,q = map(int,input().split())
uft = UnionFind(n)
for i in range(q):
    c,x,y = map(int,input().split())
    if c==0:
        uft.unite(x,y)
    else:
        print(1 if uft.is_same_set(x,y) else 0)