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;
pair<long long, long long> dp[20][10000][2];
long long n, k;
vector<long long> v;
long long mod = 998244353;
long long sum(long long a, long long b) {
  a %= mod;
  b %= mod;
  long long s = (a + b) % mod;
  return s;
}
long long mul(long long a, long long b) {
  a %= mod;
  b %= mod;
  long long s = (a * b) % mod;
  return s;
}
long long sz(long long n) {
  while (n) {
    v.push_back(n % 10);
    n /= 10;
  }
  reverse(v.begin(), v.end());
  return v.size();
}
pair<long long, long long> fnc(long long ind, long long mask, long long flag) {
  if (dp[ind][mask][flag].first != -1) return dp[ind][mask][flag];
  long long ans = 0, i, num, bits;
  if (ind == n) {
    long long cnt = 0;
    for (i = 0; i <= ((flag == 1) ? v[ind - 1] : 9); i++) {
      num = (mask | (1LL << i));
      bits = __builtin_popcountll(num);
      if (bits <= k) {
        ans += i;
        cnt++;
      }
    }
    ans %= mod;
    cnt %= mod;
    dp[ind][mask][flag] = {ans, cnt};
    return dp[ind][mask][flag];
  }
  long long pro = 1, cnt = 0;
  for (i = 1; i <= n - ind; i++) pro = (pro * 10) % mod;
  for (i = ((ind == 1 || mask == 0) ? 1 : 0);
       i <= ((flag == 1) ? v[ind - 1] : 9); i++) {
    num = (mask | (1LL << i));
    bits = __builtin_popcountll(num);
    if (bits <= k) {
      pair<long long, long long> p =
          fnc(ind + 1, num, ((flag == 1 && i == v[ind - 1]) ? 1 : 0));
      long long su = p.first;
      long long cn = p.second;
      if (cn != 0) ans = sum(sum(ans, mul(mul(i, pro), cn)), su);
      cnt = sum(cnt, cn);
    }
  }
  if (mask == 0) {
    pair<long long, long long> p = fnc(ind + 1, 0, 0);
    long long su = p.first;
    long long cn = p.second;
    if (cn != 0) ans = sum(ans, su);
    cnt = sum(cnt, cn);
  }
  ans %= mod;
  cnt %= mod;
  dp[ind][mask][flag] = {ans, cnt};
  return dp[ind][mask][flag];
}
signed main() {
  long long i, j, l, r;
  cin >> l >> r >> k;
  pair<long long, long long> val2, val1;
  memset(dp, -1, sizeof(dp));
  n = sz(l - 1);
  if (l != 1) val1 = fnc(1, 0, 1);
  long long io = val1.first;
  long long oh = val1.second;
  if (l == 1) io = 0;
  v.clear();
  memset(dp, -1, sizeof(dp));
  n = sz(r);
  val2 = fnc(1, 0, 1);
  long long oh2 = val2.second;
  long long io2 = val2.first;
  long long ans = (io2 - io + mod) % mod;
  cout << ans << endl;
  return 0;
}
