#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
    #next DP tech. で高速化
    DP = [0] * (6 << 10)
    N = N[::-1]

    #1桁目を埋める
    DP[((1 * 3) + 1) << 10 | 1] += 1
    now = int(N[0])
    for k in range(1, 10):
        DP[((k <= now) * 3 + 0) << 10 | 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)
        nDP = [0] * (6 << 10)
        #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
                x1 = ((nf * 3) + 1) << 10 | T
                x2 = x1 + (1 << 10)
                y1 = ((f * 3) + 0) << 10 | S
                y2, y3 = y1 + (1 << 10), y1 + (2 << 10)
                nDP[x1] += DP[y1] + DP[y3]
                nDP[x1] %= MOD
                nDP[x2] += DP[y2]
                nDP[x2] %= 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)
                    x = ((nf * 3) + 0) << 10 | T
                    y1 = ((f * 3) + 0) << 10 | S
                    y2, y3 = y1 + (1 << 10), y1 + (2 << 10)
                    nDP[x] += DP[y1] + DP[y2] + DP[y3]
                    nDP[x] %= MOD
        DP, nDP = nDP, DP
    
    ans = DP[3 << 10] + DP[((1 * 3) + 1) << 10 | 1] + DP[((1 * 3) + 2) << 10] - 1
    ans %= MOD
    return ans

print( solve(N) )