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 = [i for i in range(n+1)]
        # 木の高さを格納する（初期状態では0）
        self.rank = [0] * (n + 1)

    # 探索
    # 根ならその場号を返す
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            # 走査していく過程で親を書き換えていく（パス圧縮）
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    def union(self, x, y):
        # 根を探す
        x = self.find(x)
        y = self.find(y)

        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
    
    # 同じ集合に属するか判定
    def same(self, x, y):
        return self.find(x) == self.find(y)


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

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