You are given an undirected graph with N vertices and 0 edges. Process Q queries of the following types.

* `0 u v`: Add an edge (u, v).
* `1 u v`: Print 1 if u and v are in the same connected component, 0 otherwise.

Constraints

* 1 \leq N \leq 200,000
* 1 \leq Q \leq 200,000
* 0 \leq u_i, v_i \lt N

Input

Input is given from Standard Input in the following format:


N Q
t_1 u_1 v_1
t_2 u_2 v_2
:
t_Q u_Q v_Q


出力

For each query of the latter type, print the answer.

Example

Input

4 7
1 0 1
0 0 1
0 2 3
1 0 1
1 1 2
0 0 2
1 1 3


Output

0
1
0
1
#!/usr/bin/env python

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.par = [i for i in range(n)]
        self.rank = [0 for _ in range(n)]
        self.sizes = [1 for _ in range(n)]

    def root(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.root(self.par[x]) # 圧縮経路
            return self.par[x]
            # return root(par[x]) # 圧縮しない場合

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

    def unite(self, x, y):
        rx = self.root(x)
        ry = self.root(y)
        if rx == ry:
            return

        # par[rx] = ry # naiveな連結

        # 低い木を高い方にくっつける。
        if self.rank[rx] < self.rank[ry]:
            self.par[rx] = ry
            self.sizes[ry] += self.sizes[rx]
        else:
            self.par[ry] = rx
            self.sizes[rx] += self.sizes[ry]

        # rx木とry木の高さが同じならば、rx木を深くする。
        if self.rank[rx] == self.rank[ry]:
            self.rank[rx] += 1
             
    def size(self, x):
        '''
        xが属するグループのサイズを返す
        (連結成分の大きさ)
        '''
        return self.sizes[self.root(x)]

# main
n, q = map(int ,input().split())
uf = UnionFind(n)

for i in range(q):
    t, u, v = map(int, input().split())
    if t == 0:uf.unite(u, v)
    else:print(int(uf.same(u, v)))
