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 UnionFindTree:
    '''
    Union Find Tree
    '''

    def __init__(self):
        self.p = {}
        self.rank = {}

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

    def find(self, x):
        if x in self.p:
            if self.p[x] != x:
                # 経路圧縮
                self.p[x] = self.find(self.p[x])
            return self.p[x]
        return None

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

    def unite(self, x, y):
        self._unite(self.find(x), self.find(y))

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


n, q = map(int, input().split())
uft = UnionFindTree()
for i in range(n):
    uft.make(i)

for i in range(q):
    com, x, y = map(int, input().split())
    if com == 0:
        uft.unite(x, y)
    else:
        print(1 if uft.same(x, y) else 0)