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, n):
        self.par = [-1]*n

    def find(self, x):
        if(self.par[x] < 0):
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    def same(self, x, y):
        return self.find(x) == self.find(y)
    
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x==y:
            return False
        if self.par[x]>self.par[y]:
            x,y = y,x
        self.par[x]+=self.par[y]
        self.par[y] = x
        return True

    def size(self,x):
        return -self.par[self.find(x)]

    def count_group():
        c = 0
        for i in range(len(self.par)):
            if self.par[i]<0:
                c+=1
        return c

n,q = map(int, input().split())
uf = Unionfind(n)
for _ in range(q):
    com,x,y = map(int, input().split())
    if com==0:
        uf.union(x,y)
    else:
        print(int(uf.same(x,y)))