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 PYTHON):

Solution:
def main():
    n,q = map(int,raw_input().split())

    ds = DisjointSet(n)

    for i in xrange(q):
        com,x,y = map(int,raw_input().split())
        if com:
            ds.findSet(x)
            ds.findSet(y)
            if ds.parent[x] == ds.parent[y]:
                print 1
            else:
                print 0
        else:
            ds.findSet(x)
            ds.findSet(y)
            ds.unite(x,y)

class DisjointSet():
    def __init__(self,n):
        self.n = n
        self.parent = range(n)

    def findSet(self,x):
        if self.parent[x] == x:
            return x
        else:
            p = self.findSet(self.parent[x])
            self.parent[x] = p
            return p

    def unite(self,x,y):
        if self.parent[y] > self.parent[x]:
            self.parent[self.parent[y]] = self.parent[x]
        else:
            self.parent[self.parent[x]] = self.parent[y]

if __name__ == "__main__":
    main()