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 int D = 18, M = 998244353;
int k;
long long dp[D + 1][2][2][1 << 10], dp2[D + 1][2][1 << 10], d[D];
long long solve(long long a) {
  d[0] = a;
  for (int i = 0; i < D; ++i) {
    d[i + 1] = d[i] / 10;
    d[i] %= 10;
  }
  memset(dp, 0, sizeof(dp));
  memset(dp2, 0, sizeof(dp2));
  dp[D][0][0][0] = 1;
  for (int i = D; i; --i) {
    for (int a = 0; a < 4; ++a) {
      for (int b = 0; b < 1 << 10; ++b) {
        dp[i][a & 1][a / 2][b] %= M;
        dp2[i][a / 2][b] %= M;
      }
    }
    if (d[i - 1])
      dp[i - 1][0][1][0] += dp[i][0][0][0];
    else
      dp[i - 1][0][0][0] += dp[i][0][0][0];
    dp[i - 1][0][1][0] += dp[i][0][1][0];
    for (int j = 1; j < d[i - 1]; ++j) {
      dp[i - 1][1][1][1 << j] += dp[i][0][0][0];
      dp2[i - 1][1][1 << j] += dp[i][0][0][0] * j;
    }
    if (d[i - 1]) {
      dp[i - 1][1][0][1 << d[i - 1]] += dp[i][0][0][0];
      dp2[i - 1][0][1 << d[i - 1]] += dp[i][0][0][0] * d[i - 1];
    }
    for (int j = 1; j < 10; ++j) {
      dp[i - 1][1][1][1 << j] += dp[i][0][1][0];
      dp2[i - 1][1][1 << j] += dp[i][0][1][0] * j;
    }
    for (int b = 0; b < 1 << 10; ++b) {
      for (int j = 0; j < d[i - 1]; ++j) {
        dp[i - 1][1][1][b | 1 << j] += dp[i][1][0][b];
        dp2[i - 1][1][b | 1 << j] += dp2[i][0][b] * 10 + dp[i][1][0][b] * j;
      }
      dp[i - 1][1][0][b | 1 << d[i - 1]] += dp[i][1][0][b];
      dp2[i - 1][0][b | 1 << d[i - 1]] +=
          dp2[i][0][b] * 10 + dp[i][1][0][b] * d[i - 1];
      for (int j = 0; j < 10; ++j) {
        dp[i - 1][1][1][b | 1 << j] += dp[i][1][1][b];
        dp2[i - 1][1][b | 1 << j] += dp2[i][1][b] * 10 + dp[i][1][1][b] * j;
      }
    }
  }
  long long r = 0;
  for (int i = 0; i < 1 << 10; ++i)
    if (__builtin_popcount(i) <= k) r += dp2[0][0][i] + dp2[0][1][i];
  return r % M;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long l, r;
  cin >> l >> r >> k;
  cout << (solve(r) - solve(l - 1) + M) % M;
}
