class GfG {
 
    // function to calculate base^expo 
    // returns early if result exceeds the 
    // given limit to avoid overflow
    static int power(int base, int expo, int limit) {
        int result = 1;
        // now multiply the number we chose by itself n times
        // stop early and return if it exceeds before reaching n times even
        for (int i = 0; i < expo; i++) {
            result *= base;
 
            if (result > limit)  
                return result;
        }
        return result;
    }
 
    // function to find the 
    // n-th root of m
    // to find nth room of m means to get the number which needs to be multiplied n times so that we get m
    static int nthRoot(int n, int m) {
        // n-th root of 0 is 0
        if (m == 0) return 0;
 
        // If n is 1, the answer 
        // is m itself
        if (n == 1) return m;
 
        // binary search to find 
        // the integer root
        int low = 1, high = m;
        while (low <= high) {
            int mid = (low + high) / 2;
 
            // compute mid^n and compare it with m
            int val = power(mid, n, m);
            // choose a number in the middle of m, m/2 or whatever and 
            // do x^n to it and see if it satisties .. can we have a number that gives m after multiplied it 
            // by itself n times..
            if (val == m)
                return mid;  
            else if (val < m)
                low = mid + 1;  
            else
                high = mid - 1; 
        }
 
        return -1;
    }
 
    public static void main(String[] args) {
        int n = 3, m = 64;  
        // what can i multiply 3 times by itself to get 64.. 4 * 4 * 4
        int result = nthRoot(n, m);
        System.out.println(result); 
    }
}

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