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

buf = [i for i in range(n)]
height = [1] * n

def find(i):
    while i != buf[i]:
        i = buf[i]
    return i

def union(i, j):
    i = find(i)
    j = find(j)
    if height[i] >= height[j]:
        buf[j] = i
        height[i] = max(height[i], height[j] + 1)
    else:
        buf[i] = j
        height[j] = max(height[j], height[i] + 1)

for _ in range(q):
    com, x, y = map(int, input().split())
    if com == 0:
        # unite                                                                                                                                                                                             
        union(x, y)
    else:
        if find(x) == find(y):
            print(1)
        else:
            print(0)