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:
n, q = map(int, input().split())

# union find
# 最初は全て値であるとして初期化する parents[i] = i

# root(x) => xの根を返す関数
# unit(x,y) => root(x)とroot(y)が同じであればそのまま。違う場合は、xの根であるroot(x)をyの根であるroot(y)につける. 根が同じでない二つの木を合併する。
# same(x,y) => 二つのデータx,yが属する木が同じ場合、trueを返す。if root(x) == root(y)するだけ

parents = [i for i in range(n)]


def root(x):
    if parents[x] == x:
        return x
    parents[x] = root(parents[x])
    return parents[x]


def unit(x, y):
    root_x = root(x)
    root_y = root(y)
    if root_x < root_y:
        parents[root_y] = root_x
    else:
        parents[root_x] = root_y


def same(x, y):
    root_x = root(x)
    root_y = root(y)
    return (root_x == root_y)


for _ in range(q):
    query, x, y = map(int, input().split())
    if query == 0:
        unit(x, y)
    else:
        if same(x, y):
            print(1)
        else:
            print(0)