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木の構築
par=[-1]*n

def find(v):
    if par[v]<0:
        return v
    else:
        par[v]=find(par[v])
        return par[v]

def union(u,v):
    root_u=find(u)
    root_v=find(v)

    if root_u==root_v:
        return 
    
    #大きい方に接続
    if par[root_u]>par[root_v]:
        par[root_v]+=par[root_u]
        par[root_u]=root_v
    else:
        par[root_u]+=par[root_v]
        par[root_v]=root_u

def same(u,v):
    if find(u)==find(v):
        return True
    else:
        return False

for i in range(q):
    c,x,y=map(int,input().split())
    if c==0:
        union(x,y)
    else:
        ans=same(x,y)
        if ans:
            print(1)
        else:
            print(0)