function answerQuery(l: number, r: number): number {
    let memo_factorial = new Map<bigint, bigint>([[0n, 1n], [1n, 1n]])
    function factorial(n: bigint): bigint {
        if (!memo_factorial.has(n)) {
            let i_ = Array.from(memo_factorial.keys()).pop() || 1n;
            let r_ = memo_factorial.get(i_)!
            for (let i = i_ + 1n; i <= n; i++) {
                r_ *= i
                memo_factorial.set(i, r_)
            }
        }
        return memo_factorial.get(n)!;
    }

    // 0. get the substring from [l] and [r], get [frequency]
    const s = str.substring(l - 1, r)
    const frequency = new Map<string, number>()
    for (let c of s) frequency.set(c, (frequency.get(c) || 0) + 1)

    /**
     *              factorial(p_alls)
     *          m * ------------------------------- (modulo 10^9 + 7)
     *              factorial(p*...)
     * 
     * with:
     *  + [m]: numbers of character that odd, mean it can be middle
     *  + [p]: colection of character that can be appears in 2 side of palidrome
     */

    /**
     * Update 1: Number exceeded Number.MAX_SAFE_INTEGER -> using BigInt
     * Update 2: Maximum call stack exceeded -> changes [factorial] to use for...loop insteads of recursive
     * Update 3: Some test timeout -> try memo
     * Update 4: Some test with 100.000 has runtime error, idk where cause 50kb limit custom test
     */
    
    // 1. calculate prepare for formular
    let p_alls = factorial(BigInt(Array.from(frequency.values()).reduce((_, v) => _ + Math.floor(v / 2), 0)))
    let p_each = Array.from(frequency.values()).reduce((_, v) => _ * factorial(BigInt(Math.floor(v / 2))), 1n)
    let m = BigInt(Math.max(Array.from(frequency.values()).reduce((_, v) => _ + v % 2, 0), 1))
    
    // 2. apply formular
    return Number((m * p_alls / p_each) % BigInt(10 ** 9 + 7))
}