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:
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_1_A&lang=ja

# union-find

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

# xの根を求める
def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])
        return par[x]
#xとyの属する集合の併合
def unite(x,y):
    x = find(x)
    y = find(y)
    
    if x == y:
        return False
    else:
        #sizeの大きいほうがx
        if par[x] > par[y]:
            x,y = y,x
        par[x] += par[y]
        par[y] = x
        return True

#xとyが同じ集合に属するかの判定
def same(x,y):
    return find(x) == find(y)

#xが属する集合の個数
def size(x):
    return -par[find(x)]

#初期化
#根なら-size,子なら親の頂点
par = [-1] * n


for _ in range(q):
    com, ai, bi = map(int, input().split())
    # ai, bi = ai - 1, bi - 1

    # com = 0 (unite) / 1 (same)
    if com == 0:
        unite(ai, bi)
    else:
        print(1 if same(ai, bi) else 0)