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 d_all_your_paths_are_different_lengths(L):
    # l=2から始めて、lを2倍するか+1するかを何回か行ってLにしたい。
    # その回数を最小にするには、Lから始めて、Lが偶数なら2で割って奇数なら1引くことを
    # 繰り返した回数を調べればいい。
    do = []  # 2からLにするには*2と+1をどのように行えばいいか？
    while L > 2:
        if L % 2 == 0:
            do.append('*2')
            L >>= 1
        else:
            do.append('+1')
            L -= 1
    do.reverse()

    # 解説(英語版の Solution 2.)に従って辺を構成
    vertex = 2  # 現時点での頂点数
    edge = [[1, 2, 0], [1, 2, 1]]  # L=2のときはこのようにつなぐ他ない
    l = 2
    for op in do:
        if op == '*2':
            # 各辺の長さを2倍、新しい頂点に長さ0,1の辺をつなぐ
            edge = [[v1, v2, c * 2] for v1, v2, c in edge]
            edge.append([vertex, vertex + 1, 0])
            edge.append([vertex, vertex + 1, 1])
            vertex += 1
            l *= 2
        elif op == '+1':
            # 先頭と末尾の頂点に長さlの辺をつなぐ
            edge.append([1, vertex, l])
            l += 1

    # 出力形式に沿った文字列を構成
    ans = '{} {}\n'.format(vertex, len(edge))
    ans += '\n'.join([' '.join(map(str, row)) for row in edge])
    return ans

L = int(input())
print(d_all_your_paths_are_different_lengths(L))