Count Good Numbers LeetCode solution in Java

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class Solution {
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  int mod = (int) 1e9 + 7;
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  public int countGoodNumbers(long n) {
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    long first =
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        (n % 2 == 0 ? (n / 2) : (n / 2) + 1); // deciding n/2 or n/2+1 depending on n is even or odd
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    long second = n / 2; // second power would be n/2 only irrespective of even or odd
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    long mul1 = power(5, first) % mod; // 5 power n/2
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    long mul2 = power(4, second) % mod; // 4 power n/2
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    long ans = 1;
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    ans = (ans * mul1) % mod; // computing total product
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    ans = (second != 0) ? (ans * mul2) % mod : ans; // computing total product
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    return (int) (ans % mod);
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  }
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  public long power(
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      long x, long y) { // this method computes pow(x,y) in O(logn) using divide & conquer
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    long temp;
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    if (y == 0) return 1; // base case (x power 0 = 1)
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    temp = power(x, y / 2); // computing power for pow(x,y/2) -> divide & conquer step
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    if (y % 2 == 0)
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      return (temp * temp)
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          % mod; // using that result of subproblem (2 power 2 = 2 power 1 * 2 power 1)
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    else
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      return (x * temp * temp)
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          % mod; // using that result of subproblem (2 power 3 = 2 power 1 * 2 power 1 * 2)
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    // if y is odd, x power y = x power y/2 * x power y/2 * x
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    // if y is even, x power y = x power y/2 * x power y/2
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  }
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}