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 DisjointSet():
    def __init__(self, size):
        self.rank = [None] * size
        self.p = [None] * size
        for i in range(size):
            self.makeSet(i)

    def makeSet(self, x):
        self.p[x] = x
        self.rank[x] = 0

    def same(self,x,y):
        return self.findSet(x) == self.findSet(y)

    def unite(self,x,y):
        self.link(self.findSet(x), self.findSet(y))

    def link(self, x, y):
        if self.rank[x] > self.rank[y]:
            self.p[y] = x
        else:
            self.p[x] = y
            if self.rank[x] == self.rank[y]:
                self.rank[y] += 1

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

n,q = map(int, input().split())
ds = DisjointSet(n)
for i in range(q):
    t,a,b = map(int, input().split())
    if t == 0:
        ds.unite(a,b)
    elif t == 1:
        if ds.same(a,b):
            print("1")
        else:
            print("0")