You are given two integers l and r (l ≤ r). Your task is to calculate the sum of numbers from l to r (including l and r) such that each number contains at most k different digits, and print this sum modulo 998244353.

For example, if k = 1 then you have to calculate all numbers from l to r such that each number is formed using only one digit. For l = 10, r = 50 the answer is 11 + 22 + 33 + 44 = 110.

Input

The only line of the input contains three integers l, r and k (1 ≤ l ≤ r < 10^{18}, 1 ≤ k ≤ 10) — the borders of the segment and the maximum number of different digits.

Output

Print one integer — the sum of numbers from l to r such that each number contains at most k different digits, modulo 998244353.

Examples

Input

10 50 2


Output

1230


Input

1 2345 10


Output

2750685


Input

101 154 2


Output

2189

Note

For the first example the answer is just the sum of numbers from l to r which equals to (50 ⋅ 51)/(2) - (9 ⋅ 10)/(2) = 1230. This example also explained in the problem statement but for k = 1.

For the second example the answer is just the sum of numbers from l to r which equals to (2345 ⋅ 2346)/(2) = 2750685.

For the third example the answer is 101 + 110 + 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 119 + 121 + 122 + 131 + 133 + 141 + 144 + 151 = 2189.
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long l, r;
int k, lenl, lenr;
int a[20], b[20];
pair<long long, long long> dp[20][(1 << 10)][2];
long long all = 0, Pow[25];
pair<long long, long long> dfsr(int pos, int mx, int s, bool x) {
  if (pos <= 0) return (pair<long long, long long>){1, 0};
  if (!mx && ~dp[pos][s][x].second) return dp[pos][s][x];
  pair<long long, long long> ans = (pair<long long, long long>){0, 0};
  int up = mx ? a[pos] : 9;
  for (int i = 0; i <= up; i++) {
    pair<long long, long long> t;
    if (x == 0 && !i)
      t = dfsr(pos - 1, mx && (i == up), 0, 0);
    else {
      int ss = s | ((x || i) << i);
      if (__builtin_popcount(ss) > k) continue;
      t = dfsr(pos - 1, mx && (i == up), ss, 1);
    }
    ans.first = (ans.first + t.first) % mod;
    ans.second =
        (ans.second + t.second + 1ll * i * Pow[pos - 1] % mod * t.first % mod) %
        mod;
  }
  if (!mx) dp[pos][s][x] = ans;
  return ans;
}
pair<long long, long long> dfsl(int pos, int mx, int s, bool x) {
  if (pos <= 0) return (pair<long long, long long>){1, 0};
  if (!mx && ~dp[pos][s][x].second) return dp[pos][s][x];
  pair<long long, long long> ans = (pair<long long, long long>){0, 0};
  int up = mx ? b[pos] : 9;
  for (int i = 0; i <= up; i++) {
    pair<long long, long long> t;
    if (x == 0 && !i)
      t = dfsl(pos - 1, mx && (i == up), 0, 0);
    else {
      int ss = s | ((x || i) << i);
      if (__builtin_popcount(ss) > k) continue;
      t = dfsl(pos - 1, mx && (i == up), ss, 1);
    }
    ans.first = (ans.first + t.first) % mod;
    ans.second =
        (ans.second + t.second + 1ll * i * Pow[pos - 1] % mod * t.first % mod) %
        mod;
  }
  if (!mx) dp[pos][s][x] = ans;
  return ans;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%lld%lld%d", &l, &r, &k);
  Pow[0] = 1;
  for (int i = 1; i <= 20; i++) Pow[i] = Pow[i - 1] * 10 % mod;
  while (r) {
    a[++lenr] = r % 10;
    r /= 10;
  }
  l--;
  while (l) {
    b[++lenl] = l % 10;
    l /= 10;
  }
  long long ans = dfsr(lenr, 1, 0, 0).second;
  memset(dp, -1, sizeof(dp));
  ans = (ans - dfsl(lenl, 1, 0, 0).second + mod) % mod;
  printf("%lld\n", ans);
}
