#yukicoder 2772 Appearing Even Times

N = input()
MOD = 998244353

def brute(N: str):
    n = int(N)
    ans = 0
    for i in range(1, n + 1):
        i = str(i)
        C = [0] * 10
        for j in i:
            C[ int(j) ] ^= 1
        if not any(C):
            ans += 1
    return ans

def solve(N: str):
    #DP[i][f][g][S]: Nの下からi桁目まで見たとき、0から9までの数字の出現回数フラグがS、
    #                f = 以下フラグ
    #                g = 0, 1, 2: leading zerosがない、0が奇数個/偶数個つながる　状態数
    DP = [[[[0] * (1 << 10) for g in range(3)] for f in range(2)] for i in range(len(N))]
    has_bit = lambda S, x: S >> x & 1
    N = N[::-1]

    #1桁目を埋める
    DP[0][1][1][1] = 1
    now = int(N[0])
    for k in range(1, 10):
        DP[0][k <= now][0][1 << k] += 1

    #2桁目以降を決定する
    for i, now in enumerate(N[1:], start = 1):
        now = int(now)
        #0をつなげる場合
        for S in range(1 << 10):
            for f in range(2):
                nf = (now > 0) | f
                T = S ^ 1
                DP[i][nf][1][T] += DP[i - 1][f][0][S] + DP[i - 1][f][2][S]
                DP[i][nf][1][T] %= MOD
                DP[i][nf][2][T] += DP[i - 1][f][1][S]
                DP[i][nf][2][T] %= MOD

        #1 - 9をつなげる場合
        for S in range(1 << 10):
            for k in range(1, 10):
                T = S ^ (1 << k)
                for f in range(2):
                    nf = (now > k) | ((now == k) & f)
                    DP[i][nf][0][T] += sum(DP[i - 1][f][g][S] for g in range(3))
                    DP[i][nf][0][T] %= MOD
    ans = DP[-1][1][0][0] + DP[-1][1][1][1] + DP[-1][1][2][0] - 1
    ans %= MOD
    return ans

print( solve(N) )