## 解法

$O(N)$。 $2$進数でやると$N \le 20$なので足りないが、$3$進数でやると上手く収まる。 辺重み$w \le 10^6$という制約に注意。

## 実装

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i))
using namespace std;

constexpr int N = 20;
constexpr int MAX_L = 1000000;
constexpr int K = 13;
vector<tuple<int, int, int> > solve(int l) {
vector<int> e(K + 1);
e[0] = 1;
REP (k, K) e[k + 1] = 3 * e[k];
assert (e[K - 1] < MAX_L and MAX_L < e[K]);

assert (l <= MAX_L);
vector<int> trits;
for (int acc = l; acc; acc /= 3) {
trits.push_back(acc % 3);
}

vector<tuple<int, int, int> > edges;
REP (k, (int)trits.size() - 1) {
edges.emplace_back(N - k - 2, N - k - 1, 0);
edges.emplace_back(N - k - 2, N - k - 1, e[k]);
edges.emplace_back(N - k - 2, N - k - 1, 2 * e[k]);
}
int acc = 0;
REP_R (k, trits.size()) {
if (trits[k] == 1) {
edges.emplace_back(0, N - k - 1, acc);
} else if (trits[k] == 2) {
edges.emplace_back(0, N - k - 1, acc);
edges.emplace_back(0, N - k - 1, acc + e[k]);
}
acc += trits[k] * e[k];
}

array<bitset<MAX_L>, N> dp = {};
dp[0][0] = true;
REP (i, N - 1) {
for (auto edge : edges) {
int u, v, w; tie(u, v, w) = edge;
if (u != i) continue;
assert (dp[u].count() == (dp[u] << w).count());
assert ((dp[v] & (dp[u] << w)).none());
dp[v] |= dp[u] << w;
}
}
assert (dp[N - 1].count() == l and (dp[N - 1] >> l).none());
assert (edges.size() <= 60);
return edges;
}

int main() {
int l; cin >> l;
auto edges = solve(l);
cout << N << ' ' << edges.size() << endl;
for (auto edge : edges) {
int u, v, w; tie(u, v, w) = edge;
cout << u + 1 << ' ' << v + 1 << ' ' << w << endl;
}
return 0;
}