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 Node():
    def __init__(self, id):
        self.id = id
        self.p = self
        self.rank = 0


class DisjointSet():
    def __init__(self, n):
        self.nodes = {}
        for i in range(n):
            self.makeSet(i)

    def makeSet(self, i):
        self.nodes[i] = Node(i)

    def findSet(self, x):
        if x.p is not x:
            x.p = self.findSet(x.p)
        return x.p

    def unite(self, x, y):
        # ランクの低い方を子にする
        px = self.findSet(x)
        py = self.findSet(y)
        if px.rank > py.rank:
            py.p = px
        else:
            px.p = py
        if x.rank == y.rank:
            py.rank += 1

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


n, q = map(int, input().split())
uf = DisjointSet(n)
for i in range(q):
    com, ix, iy = map(int, input().split())
    if com == 0:
        # unite
        uf.unite(uf.nodes[ix], uf.nodes[iy])
    else:
        # same
        print(1 if uf.bool(uf.nodes[ix], uf.nodes[iy]) else 0)