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 k, i, j, x, dp[21][1024][2], sum[21][1024][2];
long long f(long long x) { return (1LL << x); }
long long calc(long long x) {
  vector<long long> now;
  do {
    now.push_back(x % 10);
    x /= 10;
  } while (x > 0);
  reverse(now.begin(), now.end());
  long long ans = 0;
  for (long long SZ = 1; SZ <= now.size(); ++SZ) {
    for (i = 1; i <= SZ; ++i)
      for (j = 1; j < 1024; ++j)
        for (x = 0; x < 2; ++x) dp[i][j][x] = 0, sum[i][j][x] = 0;
    if (SZ == now.size()) {
      for (i = 1; i <= 9; ++i) {
        if (i < now[0]) dp[1][f(i)][0] = 1, sum[1][f(i)][0] = i;
        if (i == now[0]) dp[1][f(i)][1] = 1, sum[1][f(i)][1] = i;
      }
    } else {
      for (i = 1; i <= 9; ++i) {
        dp[1][f(i)][0] = 1;
        sum[1][f(i)][0] = i;
      }
    }
    for (i = 1; i < SZ; ++i) {
      for (j = 1; j < 1024; ++j) {
        for (x = 0; x <= 9; ++x) {
          long long mask = j | f(x);
          if (__builtin_popcount(mask) <= k) {
            dp[i + 1][mask][0] = (dp[i][j][0] + dp[i + 1][mask][0]) % mod;
            sum[i + 1][mask][0] =
                (sum[i][j][0] * 10LL + sum[i + 1][mask][0] + dp[i][j][0] * x) %
                mod;
            if (x < now[i]) {
              dp[i + 1][mask][0] = (dp[i][j][1] + dp[i + 1][mask][0]) % mod;
              sum[i + 1][mask][0] = (sum[i][j][1] * 10LL + sum[i + 1][mask][0] +
                                     dp[i][j][1] * x) %
                                    mod;
            }
            if (x == now[i]) {
              dp[i + 1][mask][1] = (dp[i][j][1] + dp[i + 1][mask][1]) % mod;
              sum[i + 1][mask][1] = (sum[i][j][1] * 10LL + sum[i + 1][mask][1] +
                                     dp[i][j][1] * x) %
                                    mod;
            }
          }
        }
      }
    }
    for (i = 1; i <= 1024; ++i)
      for (j = 0; j < 2; ++j) ans = (ans + sum[SZ][i][j]) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long l, r;
  cin >> l >> r >> k;
  long long ans = (calc(r) - calc(l - 1) + mod) % mod;
  long long real_ans = 0;
  cout << ans << endl;
}
