Problem: Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.

People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?

As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.

Input

Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.

Output

In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.

Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.

If there are multiple tests, print any of them. You can print denominations in atbitrary order.

Examples

Input

18


Output

30 4
1 5 10 25


Input

3


Output

20 2
5 2


Input

314


Output

183 4
6 5 2 139

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAX = 1e18 + 1;
const double pi = 3.1415926535897932384626433832;
const long long M = 1e9 + 7;
long long modpow(long long x, long long n, long long mod1) {
  if (n == 0) return 1 % mod1;
  long long u = modpow(x, n / 2, mod1);
  u = (u * u) % mod1;
  if (n % 2 == 1) u = (u * x) % mod1;
  return u;
}
long long POW(long long x, long long y) { return modpow(x, y, LONG_LONG_MAX); }
long long MI(long long x, long long mod = M) { return modpow(x, mod - 2, mod); }
long long gcd(long long x, long long y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
bool sumdigits(long long a) {
  long long result = 0;
  while (a > 0) {
    result += a % 10;
    a /= 10;
  }
  return result;
}
const long long Node = 2e5 + 7;
vector<long long> adj[Node];
bool vis[Node] = {false};
void addedge(long long u, long long v) {
  adj[u - 1].push_back(v - 1);
  adj[v - 1].push_back(u - 1);
}
void dfs(long long u) {
  vis[u] = true;
  for (auto x : adj[u]) {
    if (!vis[x]) {
      dfs(x);
    }
  }
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
void seive(long long prime[], long long n) {
  long long chk = 2;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i] == 0) {
      for (long long j = i * i; j <= n; j += i) prime[j] = 1;
    }
  }
}
const long long NC = 100005, MOD = M;
long long inv[NC], fac_inv[NC], fac[NC];
void initialize() {
  long long i;
  inv[1] = 1;
  for (i = 2; i <= NC - 2; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  fac[0] = fac[1] = 1;
  for (i = 2; i <= NC - 2; i++) fac[i] = i * fac[i - 1] % MOD;
  fac_inv[0] = fac_inv[1] = 1;
  for (i = 2; i <= NC - 2; i++) fac_inv[i] = inv[i] * fac_inv[i - 1] % MOD;
}
long long ncr(long long n, long long r) {
  if (n < r) return 0;
  return (fac[n] * fac_inv[r] % MOD) * fac_inv[n - r] % MOD;
}
void solve() {
  long long n;
  cin >> n;
  cout << 2 * n - 1 << " " << 2 << endl;
  cout << 1 << " " << 2 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}