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:
    __slots__ = ["forest"]

    def __init__(self, forest: list):
        self.forest = forest

    def get_root(self, x: int) -> int:
        if self.forest[x] < 0:
            return x
        else:
            self.forest[x] = self.get_root(self.forest[x])
            return self.forest[x]

    def unite(self, x, y) -> None:
        root_x, root_y = self.get_root(x), self.get_root(y)
        if root_x != root_y:
            bigroot, smallroot = \
                (root_x, root_y) if self.forest[root_x] < self.forest[root_y]\
                else (root_y, root_x)
            self.forest[bigroot] += self.forest[smallroot]
            self.forest[smallroot] = bigroot

n, q = map(int, input().split())
uf = UnionFind([-1 for _ in [None]*n])
for _ in [None]*q:
    c, x, y = map(int, input().split())
    if c == 0:
        uf.unite(x, y)
    else:
        print(1 if uf.get_root(x) == uf.get_root(y) else 0)