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 INF = 1e18, MOD = 998244353;
const long double EPS = 1e-7;
void xadd(pair<long long, long long>& dp1, pair<long long, long long> dp2,
          long long delta) {
  dp1.first += dp2.first;
  dp1.first %= MOD;
  dp1.second += 10 * dp2.second;
  dp1.second %= MOD;
  dp1.second += delta * dp2.first;
  dp1.second %= MOD;
}
long long get_new_mask(long long mask, long long new_dig) {
  if (mask == 0 && new_dig == 0) {
    return 0;
  }
  return mask | (1 << new_dig);
}
long long get_ans(long long n, long long k) {
  if (n == 0) {
    return 0;
  }
  vector<long long> dig;
  while (n != 0) {
    dig.push_back(n % 10);
    n /= 10;
  }
  reverse(dig.begin(), dig.end());
  pair<long long, long long> dp[(signed)dig.size()][1 << 10][2];
  for (long long i = 0; i < (signed)dig.size(); ++i) {
    for (long long mask = 0; mask < (1 << 10); ++mask) {
      dp[i][mask][0] = dp[i][mask][1] = {0, 0};
    }
  }
  for (long long i = 0; i < dig[0]; ++i) {
    dp[0][get_new_mask(0, i)][0] = {1, i};
  }
  dp[0][get_new_mask(0, dig[0])][1] = {1, dig[0]};
  for (long long i = 0; i + 1 < (signed)dig.size(); ++i) {
    for (long long mask = 0; mask < (1 << 10); ++mask) {
      for (long long j = 0; j < dig[i + 1]; ++j) {
        long long new_mask = get_new_mask(mask, j);
        if (__builtin_popcount(new_mask) <= k) {
          xadd(dp[i + 1][new_mask][0], dp[i][mask][0], j);
          xadd(dp[i + 1][new_mask][0], dp[i][mask][1], j);
        }
      }
      long long new_mask1 = get_new_mask(mask, dig[i + 1]);
      if (__builtin_popcount(new_mask1) <= k) {
        xadd(dp[i + 1][new_mask1][1], dp[i][mask][1], dig[i + 1]);
      }
      for (long long j = dig[i + 1]; j < 10; ++j) {
        long long new_mask = get_new_mask(mask, j);
        if (__builtin_popcount(new_mask) <= k) {
          xadd(dp[i + 1][new_mask][0], dp[i][mask][0], j);
        }
      }
    }
  }
  long long res = 0;
  for (long long mask = 0; mask < (1 << 10); ++mask) {
    res += dp[(signed)dig.size() - 1][mask][0].second;
    res %= MOD;
    res += dp[(signed)dig.size() - 1][mask][1].second;
    res %= MOD;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long l, r, k;
  cin >> l >> r >> k;
  cout << (get_ans(r, k) - get_ans(l - 1, k) + MOD) % MOD << endl;
  return 0;
}
