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;
long long l, r;
int k;
const long long mod = 998244353;
const int Mx = 1 << 10;
long long dp[20][Mx][2];
long long dp2[20][Mx][2];
int pop_count(long long x) {
  int res = 0;
  while (x) {
    res += (x & 1);
    x >>= 1;
  }
  return res;
}
long long solve(long long n) {
  memset(dp, 0, sizeof(dp));
  memset(dp2, 0, sizeof(dp2));
  int len = 0;
  long long tmp = n;
  int digits[20];
  while (tmp) {
    digits[len] = tmp % 10;
    tmp /= 10;
    len++;
  }
  for (int i = 0; i < 10; i++) {
    dp[0][1 << i][0] = 1;
    dp2[0][1 << i][0] = i;
    if (i <= digits[0]) {
      dp[0][1 << i][1] = 1;
      dp2[0][1 << i][1] = i;
    }
  }
  long long val = 1;
  for (int pos = 1; pos < len; pos++) {
    val *= 10;
    val %= mod;
    for (int mask = 1; mask < Mx; mask++) {
      int s = (pos == len - 1) ? 1 : 0;
      int e = (pos == len - 1) ? digits[pos] + 1 : 10;
      for (int i = s; i < e; i++) {
        int new_mask = mask | (1 << i);
        if (pop_count(new_mask) > k) continue;
        long long tmp = (val * i) % mod;
        dp[pos][new_mask][0] += dp[pos - 1][mask][0];
        dp[pos][new_mask][0] %= mod;
        dp2[pos][new_mask][0] +=
            tmp * dp[pos - 1][mask][0] + dp2[pos - 1][mask][0];
        dp2[pos][new_mask][0] %= mod;
        if (i < digits[pos]) {
          dp[pos][new_mask][1] += dp[pos - 1][mask][0];
          dp[pos][new_mask][1] %= mod;
          dp2[pos][new_mask][1] +=
              tmp * dp[pos - 1][mask][0] + dp2[pos - 1][mask][0];
          dp2[pos][new_mask][1] %= mod;
        } else if (i == digits[pos]) {
          dp[pos][new_mask][1] += dp[pos - 1][mask][1];
          dp[pos][new_mask][1] %= mod;
          dp2[pos][new_mask][1] +=
              tmp * dp[pos - 1][mask][1] + dp2[pos - 1][mask][1];
          dp2[pos][new_mask][1] %= mod;
        }
      }
    }
  }
  long long res = 0;
  for (int mask = 1; mask < Mx; mask++) {
    if (len > 0) res += dp2[len - 1][mask][1];
    res %= mod;
  }
  return res;
}
long long calc(long long n) {
  long long tmp = 0;
  long long res = 0;
  while (1) {
    tmp *= 10;
    tmp += 9;
    if (tmp >= n) {
      res += solve(n);
      res %= mod;
      break;
    }
    res += solve(tmp);
    res %= mod;
  }
  return res;
}
int main() {
  cin >> l >> r >> k;
  long long ans = (calc(r) - calc(l - 1) + mod) % mod;
  cout << ans << endl;
  return 0;
}
