Challenge: Fast Power (mod)

Lab C++ Beginner 10 points

Read a, b and m. Print a^b mod m using a recursive function that halves b each time. Take the remainder after every multiplication so numbers stay small.

Input

Three whole numbers a b m (0 ≤ a, b ≤ 10¹⁸, 1 ≤ m ≤ 10⁹).

Output

One number.

Example

Input

2 10 1000

Output

Sample input available Sample output available

Editor

C++

Hints

Base case: b == 0 gives 1 (mod m).

half = powmod(a, b / 2, m); square it, then multiply by a if b is odd.

Use % m after every multiplication.

Solution explanation

About 60 calls even for b = 10¹⁸. Because every value stays below m ≤ 10⁹, products fit in a long long.

Show solution

#include <iostream>
using namespace std;

long long powmod(long long a, long long b, long long m) {
    if (b == 0) return 1 % m;
    long long half = powmod(a, b / 2, m);
    long long result = half * half % m;
    if (b % 2 == 1) result = result * (a % m) % m;
    return result;
}

int main() {
    long long a, b, m;
    cin >> a >> b >> m;
    cout << powmod(a, b, m) << endl;
    return 0;
}

Output

Answer