#include <bits/stdc++.h>
using namespace std;
long long Chinese_Remainder_Theorem();
long long ans = 0;
long long power(long long a, long long b);
long long gcd(long long a, long long b, long long &x, long long &y);
long long fib(long long n);
bool is_prime(long long n);
vector<long long> prime_factors(long long n);
vector<long long> sieve(long long n);
long long euler_totient(long long n);
void Matrix_Power(long long A[2][2], long long n);
void multiply(long long A[2][2], long long B[2][2]);
int main() {
  ios_base::sync_with_stdio(false);
  std::cout << std::fixed;
  std::cout << std::setprecision(6);
  double f1;
  cin >> f1;
  double store = (44 / 7) * f1;
  cout << f1 << "\n";
  return 0;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long half = power(a, b / 2);
  long long ret = (half * half) % 1000000007;
  if (b % 2 == 1) ret = (ret * a) % 1000000007;
  return ret;
}
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  ans = (x + b) % b;
  return d;
}
long long fib(long long n) {
  if (n == 0) return 0;
  long long A[2][2] = {{1, 1}, {1, 0}};
  Matrix_Power(A, n - 1);
  return A[0][0];
}
void Matrix_Power(long long A[2][2], long long n) {
  if (n == 0 || n == 1) {
    return;
  }
  Matrix_Power(A, n / 2);
  multiply(A, A);
  if (n % 2 == 1) {
    long long B[2][2] = {{1, 1}, {1, 0}};
    multiply(A, B);
  }
}
void multiply(long long A[2][2], long long B[2][2]) {
  long long x, y, z, w;
  x = (A[0][0] * B[0][0]) + (A[0][1] * B[1][0]);
  y = (A[0][0] * B[0][1]) + (A[0][1] * B[1][1]);
  z = (A[1][0] * B[0][0]) + (A[1][1] * B[1][0]);
  w = (A[1][0] * B[0][1]) + (A[1][1] * B[1][1]);
  A[0][0] = x;
  A[0][1] = y;
  A[1][0] = z;
  A[1][1] = w;
}
bool is_prime(long long n) {
  for (long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
vector<long long> prime_factors(long long n) {
  vector<long long> v1;
  while (n % 2 == 0) {
    v1.push_back(2);
    n /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n == 1) break;
    while (n % i == 0) {
      v1.push_back(i);
      n /= i;
    }
  }
  if (n > 1) v1.push_back(n);
  return v1;
}
vector<long long> sieve(long long n) {
  vector<long long> v1(n + 1);
  long long i, j;
  v1[0] = v1[1] = 0;
  for (i = 2; i <= n; i++) {
    v1[i] = i;
  }
  for (i = 2; i <= n; i++) {
    if (v1[i] == 0) continue;
    for (j = i + i; j <= n; j += i) {
      v1[j] = 0;
    }
  }
  vector<long long> v2;
  for (i = 1; i <= n; i++) {
    if (v1[i] != 0) {
      v2.push_back(i);
    }
  }
  return v2;
}
long long euler_totient(long long n) {
  if (n == 1) return 1;
  vector<long long> v1 = prime_factors(n);
  long long i;
  vector<long long> v2;
  v2.push_back(v1[0]);
  for (i = 1; i < v1.size(); i++) {
    if (v1[i] == v1[i - 1]) continue;
    v2.push_back(v1[i]);
  }
  long long store = n;
  for (i = 0; i < v2.size(); i++) {
    store *= (v2[i] - 1);
  }
  for (i = 0; i < v2.size(); i++) {
    store /= v2[i];
  }
  return store;
}
long long Chinese_Remainder_Theorem() {
  cout << "ENTER THE NUMBER OF EQUATIONS\n";
  long long n, i, num1, num2;
  cin >> n;
  pair<long long, long long> p1[n + 1];
  for (i = 1; i <= n; i++) {
    cin >> num1 >> num2;
    p1[i] = make_pair(num1, num2);
  }
  vector<long long> v1(n + 1);
  long long M = 1;
  for (i = 1; i <= n; i++) {
    M *= p1[i].second;
  }
  for (i = 1; i <= n; i++) {
    v1[i] = M / p1[i].second;
  }
  long long X, sum = 0;
  long long a, b, x, y;
  for (i = 1; i <= n; i++) {
    a = v1[i];
    b = p1[i].second;
    gcd(a, b, x, y);
    x = p1[i].first * v1[i] * ans;
    sum += x % M;
  }
  return sum % M;
}
