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 N = 20;
const int M = 1 << 10;
const int MOD = 998244353;
int k;
int pw10[N];
int bitCnt[M];
pair<long long, int> dp[N][M][2];
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  if (a < 0) a += MOD;
  return a;
}
int mul(int a, int b) { return a * 1ll * b % MOD; }
int calc(const string &s) {
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      dp[i][j][0] = dp[i][j][1] = {0ll, 0};
    }
  }
  int len = s.size();
  dp[0][0][1] = {1ll, 0};
  for (int i = 0; i < len; ++i) {
    int cdig = s[i] - '0';
    for (int mask = 0; mask < M; ++mask) {
      if (dp[i][mask][0].first == 0 && dp[i][mask][1].first == 0) continue;
      for (int dig = (i == 0); dig <= 9; ++dig) {
        int nmask = mask | (1 << dig);
        long long cnt = dp[i][mask][0].first;
        int sum = add(dp[i][mask][0].second,
                      mul(dig, mul(pw10[len - i - 1], cnt % MOD)));
        dp[i + 1][nmask][0].first += cnt;
        dp[i + 1][nmask][0].second = add(dp[i + 1][nmask][0].second, sum);
      }
      for (int dig = (i == 0); dig <= cdig; ++dig) {
        int nmask = mask | (1 << dig);
        long long cnt = dp[i][mask][1].first;
        int sum = add(dp[i][mask][1].second,
                      mul(dig, mul(pw10[len - i - 1], cnt % MOD)));
        dp[i + 1][nmask][dig == cdig].first += cnt;
        dp[i + 1][nmask][dig == cdig].second =
            add(dp[i + 1][nmask][dig == cdig].second, sum);
      }
    }
  }
  int ans = 0;
  for (int mask = 0; mask < M; ++mask) {
    if (bitCnt[mask] > k) continue;
    ans = add(ans, dp[len][mask][0].second);
    ans = add(ans, dp[len][mask][1].second);
  }
  return ans;
}
int calc(long long n) {
  int len = to_string(n).size();
  int ans = 0;
  for (int l = 1; l < len; ++l) {
    ans = add(ans, calc(string(l, '9')));
  }
  ans = add(ans, calc(to_string(n)));
  return ans;
}
int main() {
  pw10[0] = 1;
  for (int i = 1; i < N; ++i) {
    pw10[i] = mul(pw10[i - 1], 10);
  }
  for (int i = 0; i < M; ++i) {
    bitCnt[i] = __builtin_popcount(i);
  }
  long long l, r;
  cin >> l >> r >> k;
  int ans = add(calc(r), -calc(l - 1));
  cout << ans << endl;
  return 0;
}
