/* -*- coding: utf-8 -*-
 *
 * 2772.cc:  No.2772 Appearing Even Times - yukicoder
 */

#include<cstdio>
#include<cstring>
#include<algorithm>
 
using namespace std;

/* constant */

const int MAX_N = 10001;
const int M = 10;
const int MBITS = 1 << M;
const int MOD = 998244353;

/* typedef */

template<const int MOD>
struct MI {
  int v;
  MI(): v() {}
  MI(int _v): v(_v % MOD) { if (v < 0) v += MOD; }
  MI(long long _v): v(_v % MOD) { if (v < 0) v += MOD; }

  explicit operator int() const { return v; }
  
  MI operator+(const MI m) const { return MI(v + m.v); }
  MI operator-(const MI m) const { return MI(v + MOD - m.v); }
  MI operator*(const MI m) const { return MI((long long)v * m.v); }

  MI &operator+=(const MI m) { return (*this = *this + m); }
  MI &operator-=(const MI m) { return (*this = *this - m); }
  MI &operator*=(const MI m) { return (*this = *this * m); }

  bool operator==(const MI m) const { return v == m.v; }
  bool operator!=(const MI m) const { return v != m.v; }

  MI pow(int n) const {  // a^n % MOD
    MI pm = 1, a = *this;
    while (n > 0) {
      if (n & 1) pm *= a;
      a *= a;
      n >>= 1;
    }
    return pm;
  }

  MI inv() const { return pow(MOD - 2); }
  MI operator/(const MI m) const { return *this * m.inv(); }
  MI &operator/=(const MI m) { return (*this = *this / m); }
};

using mi = MI<MOD>;

/* global variables */

char s[MAX_N + 4];
mi dp[MAX_N + 1][MBITS], ss[MAX_N + 1];

/* subroutines */

/* main */

int main() {
  scanf("%s", s);
  int n = strlen(s);
  reverse(s, s + n);

  int co = 1;
  for (int i = 0; co && i < n; i++) {
    int d = (s[i] - '0') + co;
    s[i] = '0' + d % 10;
    co = d / 10;
  }
  if (co > 0) s[n++] = '1';
  
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int bits = 0; bits < MBITS; bits++)
      if (dp[i][bits] != 0)
	for (int j = 0; j < M; j++)
	  dp[i + 1][bits ^ (1 << j)] += dp[i][bits];
    for (int j = 1; j < M; j++)
      ss[i + 1] += dp[i][1 << j];
  }

  mi sum = 0;
  for (int i = 0; i < n; i++) sum += ss[i];

  int msk = 0;
  for (int i = n - 1; i >= 0; i--) {
    int di = s[i] - '0';
    int j0 = (i < n - 1) ? 0 : 1;
    for (int j = j0; j < di; j++) sum += dp[i][msk ^ (1 << j)];
    msk ^= (1 << di);
  }
  
  printf("%d\n", (int)sum);
  return 0;
}