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:
    MAX_N = 0
    PAR  = [] 
    RANK = []

    def __init__(self,n):
        self.MAX_N = n
        self.PAR = [ i for i in range(n)]
        self.RANK= [ 0 ] * n

    def find(self,x):
        if self.PAR[x] == x:
            return x
        else:
            self.PAR[x] = self.find(self.PAR[x])
            return self.PAR[x] 

    def union(self,x,y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if self.RANK[x] < self.RANK[y]:
            self.PAR[x] = y
        else:
            self.PAR[y] = x
            if self.RANK[x] == self.RANK[y]:
                self.RANK[x] += 1

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

N,Q  = map(int,input().split())
Ans = []
tree = UnionFind(N+1)
for _ in range(Q):
    t,u,v = map(int,input().split())
    if t == 0:
        tree.union(u,v)
    else:
        if tree.same(u,v):
            Ans.append(1)
        else:
            Ans.append(0)
for a in Ans:
    print(a)