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:
#!/usr/bin/python3

class UnionFindTree:
    def __init__(self, x):
        self.parent = self
        self.id = x
        self.rank = 0

    def find(self):
        if self.id == self.parent.id:
            return self
        else:
            return self.parent.find()


# main
n, q = map(int, input().split())
nodes = [UnionFindTree(i) for i in range(n)]
for i in range(q):
    com, x, y = map(int, input().split())
    x_root = nodes[x].find()
    y_root = nodes[y].find()
    if com == 0:
        # union
        if x_root.rank > y_root.rank:
            y_root.parent = x_root
        elif y_root.rank > x_root.rank:
            x_root.parent = y_root
        elif x_root.id != y_root.id:
            y_root.parent = x_root
            x_root.rank += 1
    else:
        # same
        same = 1 if x_root.id == y_root.id else 0
        print(same)