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, dimension):
        self.root = [i for i in range(dimension)]
        self.height = [1 for i in range(dimension)]

    def parent(self, i):
        p = self.root[i]
        if i == p:
            return p
        else:
            pp = self.parent(p)
            self.root[i] = pp
            return pp

    def unite(self, i1, i2):
        p1 = self.parent(i1)
        p2 = self.parent(i2)
        if self.height[p1] < self.height[p2]:
            self.root[p1] = p2
        else:
            self.root[p2] = p1
            if self.height[p1] == self.height[p2]:
                self.height[p1] += 1

    def same(self, i1, i2):
        return self.parent(i1) == self.parent(i2)


n, q = map(int, input().strip().split())
uf = UnionFind(n)

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