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:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

# 終了後に解説を見て解けた
# パス数が２のべき乗であれば簡単にグラフを構築できることに気付きたかった。
#
# 本番中は、10進法で考えて 頂点1と頂点2の間を辺{0, 100, 200, ..., 900}で繋ぎ、頂点2と頂点3の間を辺{0, 10, ..., 90}で繋ぎ、
# などという事を考えていたが、それ以上進展させることが難しかった。


import array
from bisect import *
from collections import *
import fractions
import heapq 
from itertools import *
import math
import random
import re
import string
import sys

L = int(input())

v = 1
edges = []  # (from, to, cost)

# L = 37の場合を例にとりコメントを付与
# まず長さ0 ~ 31までのパスを作成する。
#   頂点1 -> 頂点2を0の辺と1の辺で結ぶ
#   頂点2 -> 頂点3を0の辺と2の辺で結ぶ
#   頂点3 -> 頂点4を0の辺と4の辺で結ぶ
#   頂点4 -> 頂点5を0の辺と8の辺で結ぶ
#   頂点5 -> 頂点6を0の辺と16の辺で結ぶ
n = 2
cost = 1
while n <= L:
    edges.append((v, v+1, 0))
    edges.append((v, v+1, cost))
    cost *= 2
    v += 1
    n *= 2

# 頂点6をv_sink(最終到達点)とする
v_sink = v

# ここから先、長さ32, 33, 34, 35, 36のパスを生成することになる。
# 頂点3と頂点6を、32の辺で直でつなぐと、長さ32, 33, 34, 35のパスが作れる。
# 頂点1と頂点6を、36の辺で直でつなぐと、長さ36のパスが作れる。
n //= 2
N = n
L -= N

while L > 0:
    if L >= n:
        edges.append((v, v_sink, N))
        L -= n
        N += n
    n //= 2
    v -= 1

print(v_sink, len(edges))
for e in edges:
    print(*e)