Count Good Numbers LeetCode solution in JavaScript

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class Math1 {
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  // https://en.wikipedia.org/wiki/Modular_exponentiation
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  static modular_pow(base, exponent, modulus) {
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    if (modulus === 1n) return 0n;
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    let result = 1n;
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    base = base % modulus;
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    while (exponent > 0n) {
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      if (exponent % 2n == 1n) result = (result * base) % modulus;
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      exponent = exponent >> 1n;
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      base = (base * base) % modulus;
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    }
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    return result;
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  }
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}
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var countGoodNumbers = function (n) {
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  // NOTE: 0n, 1n, 2n, 3n, 4n, 5n are numbers in BigInt
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  n = BigInt(n); // convert to BigInt, to avoid no rounding issues
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  const odds = n / 2n,
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    evens = n - odds,
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    MOD = BigInt(Math.pow(10, 9) + 7);
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  // from wikipedia
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  return (
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    (Math1.modular_pow(4n, odds, MOD) * Math1.modular_pow(5n, evens, MOD)) % MOD
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  );
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};