Problem: You are given an integer L. Construct a directed graph that satisfies the conditions below. The graph may contain multiple edges between the same pair of vertices. It can be proved that such a graph always exists.

* The number of vertices, N, is at most 20. The vertices are given ID numbers from 1 to N.
* The number of edges, M, is at most 60. Each edge has an integer length between 0 and 10^6 (inclusive).
* Every edge is directed from the vertex with the smaller ID to the vertex with the larger ID. That is, 1,2,...,N is one possible topological order of the vertices.
* There are exactly L different paths from Vertex 1 to Vertex N. The lengths of these paths are all different, and they are integers between 0 and L-1.



Here, the length of a path is the sum of the lengths of the edges contained in that path, and two paths are considered different when the sets of the edges contained in those paths are different.

Constraints

* 2 \leq L \leq 10^6
* L is an integer.

Input

Input is given from Standard Input in the following format:


L


Output

In the first line, print N and M, the number of the vertices and edges in your graph. In the i-th of the following M lines, print three integers u_i,v_i and w_i, representing the starting vertex, the ending vertex and the length of the i-th edge. If there are multiple solutions, any of them will be accepted.

Examples

Input

4


Output

8 10
1 2 0
2 3 0
3 4 0
1 5 0
2 6 0
3 7 0
4 8 0
5 6 1
6 7 1
7 8 1


Input

5


Output

5 7
1 2 0
2 3 1
3 4 0
4 5 0
2 4 0
1 3 3
3 5 1

Solution (in PYTHON3):

Solution:
def iin(): return int(input())

l = iin()
stack = []

ans = [[1, 2, 0]]
x = l
while True:
    if x == 1:
        break
    #x > 2を条件に加えた理由: 
    #x = 1 -> 2の操作を*2で済ませると[1, 2, 0], [2, 3, 0], [2, 3, 1] となり、
    #頂点数が21以上になるケースが出るため, x = 1 -> 2は+1で済ませる
    while x % 2 == 0 and x > 2:
        x //= 2
        stack.append('*2')
    if x == 1:
        break
    x -= 1
    stack.append('+1')

last_v_num = 2
x = 1#L, つまりパスの数
while len(stack) > 0:
    op = stack.pop()
#    print(op)
    if op == '+1':
        x += 1#パスの数を増やす
        ans.append([1, last_v_num, x - 1])#頂点1と最後の頂点を直接結ぶ, 辺の重みはパスの数 - 1
    elif op == '*2':
        x *= 2#パスの数を増やす
        for i in range(len(ans)):
            ans[i][2] *= 2#w *= 2して, 既存のパスの重みを2倍にする
        #最後の頂点の後ろに, 新たな頂点を加えて結ぶ, 辺の重みは0と1
        ans.append([last_v_num, last_v_num + 1, 0])
        ans.append([last_v_num, last_v_num + 1, 1])
        last_v_num += 1

N = last_v_num
M = len(ans)
print(N, M)
for u, v, w in ans:
   print(u, v, w)