p02343 Disjoint Set: Union Find Tree
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
```python
# class UnionFind:
#
#     def __init__(self,n):
#         self.roots = range(n)
#
#     def

def root(i):
    if par[i] == i:
        return i

    par[i]=root(par[i])
    return par[i]

def unite(x,y):
    xr = root(x)
    yr = root(y)
    par[yr] = xr

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



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

par = list(range(n))

for i in range(q):
    commands = list(map(int,input().split()))
    if commands[0] == 0:
        unite(commands[1], commands[2])
    if commands[0] == 1:
        if same(commands[1], commands[2]):
            print(1)
        else:
            print(0)
```