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;
string s;
long long int k;
long long int dp[20][(1 << 10) + 5][2][2];
long long int cnt[2][20][(1 << 10) + 5][2][2];
long long int p[22];
long long int go(long long int pos, long long int mask, bool flg, bool prefix,
                 bool type) {
  if (pos == s.length()) {
    return dp[pos][mask][flg][prefix] = 0;
  }
  if (dp[pos][mask][flg][prefix] != -1) {
    return dp[pos][mask][flg][prefix];
  }
  long long int c_ans = 0;
  for (long long int d = 0; d <= 9; d++) {
    long long int c_mask = mask;
    if (!(d == 0 && prefix)) {
      c_mask |= (1ll << d);
    }
    long long int z = __builtin_popcount(c_mask);
    if (z <= k) {
      if (flg && d > (s[pos] - '0')) {
        continue;
      }
      long long int val = (d * p[s.length() - 1 - pos]) % 998244353ll;
      long long int val2 = cnt[type][pos + 1][c_mask]
                              [flg && (d == s[pos] - '0')][prefix && (d == 0)];
      val = (val * val2) % 998244353ll;
      c_ans = (c_ans + val) % 998244353ll;
      c_ans = (c_ans + go(pos + 1, c_mask, flg && (d == s[pos] - '0'),
                          prefix && (d == 0), type)) %
              998244353ll;
    }
  }
  return dp[pos][mask][flg][prefix] = c_ans;
}
long long int solve(long long int pos, long long int mask, bool flg,
                    bool prefix, bool type) {
  if (pos == s.length()) {
    return cnt[type][pos][mask][flg][prefix] = 1;
  }
  if (cnt[type][pos][mask][flg][prefix] != -1) {
    return cnt[type][pos][mask][flg][prefix];
  }
  long long int c_ans = 0;
  for (long long int d = 0; d <= 9; d++) {
    long long int c_mask = mask;
    if (!(d == 0 && prefix)) {
      c_mask |= (1ll << d);
    }
    long long int z = __builtin_popcount(c_mask);
    if (z <= k) {
      if (flg && d > (s[pos] - '0')) {
        continue;
      }
      c_ans = (c_ans + solve(pos + 1, c_mask, flg && (d == s[pos] - '0'),
                             prefix && (d == 0), type)) %
              998244353ll;
    }
  }
  return cnt[type][pos][mask][flg][prefix] = c_ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  p[0] = 1;
  for (long long int i = 1; i <= 20; i++) {
    p[i] = (p[i - 1] * 10ll) % 998244353ll;
  }
  long long int l, r;
  cin >> l >> r >> k;
  l--;
  memset(cnt, -1, sizeof(cnt));
  s = to_string(r);
  solve(0, 0, 1, 1, 0);
  s = to_string(l);
  solve(0, 0, 1, 1, 1);
  s = to_string(r);
  memset(dp, -1, sizeof(dp));
  long long int ans1 = go(0, 0, 1, 1, 0);
  s = to_string(l);
  memset(dp, -1, sizeof(dp));
  long long int ans2 = go(0, 0, 1, 1, 1);
  cout << (ans1 - ans2 + 998244353ll) % 998244353ll << "\n";
}
