#include <iostream>

/**
 * A standard practice in competitive programming for performance,
 * although not strictly necessary for a single input.
 */
void setup_io() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
}

int main() {
    setup_io();

    // The input 'n' can be up to 10^18, so we must use a 64-bit integer type.
    long long n;
    std::cin >> n;

    // --- Step 1: Calculate the minimum number of bits required to store n. ---
    // This is equivalent to finding the length of the binary representation of n.
    // For any positive integer, this can be calculated by counting how many times
    // we can divide it by 2 until it becomes 0.
    // The problem guarantees 1 <= n, so we don't need a special case for n = 0.
    int min_bits = 0;
    long long temp_n = n;
    while (temp_n > 0) {
        // Right-shifting by 1 is an efficient way to perform integer division by 2.
        temp_n >>= 1;
        min_bits++;
    }

    // --- Step 2: Find the smallest power of two 'b' such that b >= min_bits. ---
    // We can start with b = 1 and repeatedly multiply by 2 until it's large enough.
    int b = 1;
    while (b < min_bits) {
        // Left-shifting by 1 is an efficient way to multiply by 2.
        b <<= 1;
    }

    // --- Step 3: Print the final result with the correct pluralization. ---
    std::cout << b;
    if (b == 1) {
        std::cout << " bit" << std::endl;
    } else {
        std::cout << " bits" << std::endl;
    }

    return 0;
}