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, s):
        self.nodes = [-1 for _ in range(s)]

    def find(self, x):
        if self.nodes[x] < 0:
            return x
        else:
            return self.find(self.nodes[x])

    def union(self, x, y):
        s_1 = self.find(x)
        s_2 = self.find(y)
        if s_1 == s_2:
            return False
        if self.nodes[s_1] > self.nodes[s_2]:
            s_1, s_2 = s_2, s_1
        self.nodes[s_1] += self.nodes[s_2]
        self.nodes[s_2] = s_1
        return True

    def is_same(self, x, y):
        return self.find(x) == self.find(y)


if __name__ == "__main__":
    n, q = map(int, input().split())
    sets = UnionFind(n)
    for i in range(q):
        com, x, y = map(int, input().split())
        if com == 0:
            sets.union(x, y)
        else:
            result = sets.is_same(x, y)
            if result is True:
                print("1")
            else:
                print("0")