конвертировать номер в Word J

let a = [
  "",
  "one ",
  "two ",
  "three ",
  "four ",
  "five ",
  "six ",
  "seven ",
  "eight ",
  "nine ",
  "ten ",
  "eleven ",
  "twelve ",
  "thirteen ",
  "fourteen ",
  "fifteen ",
  "sixteen ",
  "seventeen ",
  "eighteen ",
  "nineteen ",
];
let b = [
  "",
  "",
  "twenty",
  "thirty",
  "forty",
  "fifty",
  "sixty",
  "seventy",
  "eighty",
  "ninety",
];

function inWords(num) {
  if ((num = num.toString()).length > 9) return "overflow";
  let n = ("000000000" + num)
    .substr(-9)
    .match(/^(\d{2})(\d{2})(\d{2})(\d{1})(\d{2})$/);
  if (!n) return;
  var str = "";
  str +=
    n[1] != 0
      ? (a[Number(n[1])] || b[n[1][0]] + " " + a[n[1][1]]) + "crore "
      : "";
  str +=
    n[2] != 0
      ? (a[Number(n[2])] || b[n[2][0]] + " " + a[n[2][1]]) + "lakh "
      : "";
  str +=
    n[3] != 0
      ? (a[Number(n[3])] || b[n[3][0]] + " " + a[n[3][1]]) + "thousand "
      : "";
  str +=
    n[4] != 0
      ? (a[Number(n[4])] || b[n[4][0]] + " " + a[n[4][1]]) + "hundred "
      : "";
  str +=
    n[5] != 0
      ? (str != "" ? "and " : "") +
        (a[Number(n[5])] || b[n[5][0]] + " " + a[n[5][1]]) +
        "only "
      : "";
  return str;
}

/* eslint eqeqeq: 0 */
sushangmi55