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>
#pragma warning(disable : 4996)
using namespace std;
using ld = long double;
template <class T>
using Table = vector<vector<T>>;
const ld eps = 1e-9;
using Graph = vector<vector<int>>;
using ll = long long;
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> v) {
  os << "( " << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<vector<T>> &v) {
  for (int i = 0; i < v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<set<T>> &v) {
  for (int i = 0; i < v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  int i = 0;
  for (auto it : v) {
    if (i > 0) {
      os << ' ';
    }
    os << it;
    i++;
  }
  return os;
}
const int mod = 998244353;
struct Mod {
 public:
  int num;
  Mod() : Mod(0) { ; }
  Mod(long long int n) : num((n % mod + mod) % mod) {
    static_assert(mod < INT_MAX / 2,
                  "mod is too big, please make num 'long long int' from 'int'");
  }
  Mod(int n) : Mod(static_cast<long long int>(n)) { ; }
  operator int() { return num; }
};
Mod operator+(const Mod a, const Mod b) { return Mod((a.num + b.num) % mod); }
Mod operator+(const long long int a, const Mod b) { return Mod(a) + b; }
Mod operator+(const Mod a, const long long int b) { return b + a; }
Mod operator++(Mod &a) { return a + Mod(1); }
Mod operator-(const Mod a, const Mod b) {
  return Mod((mod + a.num - b.num) % mod);
}
Mod operator-(const long long int a, const Mod b) { return Mod(a) - b; }
Mod operator--(Mod &a) { return a - Mod(1); }
Mod operator*(const Mod a, const Mod b) {
  return Mod(((long long)a.num * b.num) % mod);
}
Mod operator*(const long long int a, const Mod b) { return Mod(a) * b; }
Mod operator*(const Mod a, const long long int b) { return Mod(b) * a; }
Mod operator*(const Mod a, const int b) { return Mod(b) * a; }
Mod operator+=(Mod &a, const Mod b) { return a = a + b; }
Mod operator+=(long long int &a, const Mod b) { return a = a + b; }
Mod operator-=(Mod &a, const Mod b) { return a = a - b; }
Mod operator-=(long long int &a, const Mod b) { return a = a - b; }
Mod operator*=(Mod &a, const Mod b) { return a = a * b; }
Mod operator*=(long long int &a, const Mod b) { return a = a * b; }
Mod operator*=(Mod &a, const long long int &b) { return a = a * b; }
Mod operator^(const Mod a, const int n) {
  if (n == 0) return Mod(1);
  Mod res = (a * a) ^ (n / 2);
  if (n % 2) res = res * a;
  return res;
}
Mod mod_pow(const Mod a, const long long n) {
  if (n == 0) return Mod(1);
  Mod res = mod_pow((a * a), (n / 2));
  if (n % 2) res = res * a;
  return res;
}
Mod inv(const Mod a) { return a ^ (mod - 2); }
Mod operator/(const Mod a, const Mod b) {
  assert(b.num != 0);
  return a * inv(b);
}
Mod operator/(const long long int a, const Mod b) { return Mod(a) / b; }
Mod operator/=(Mod &a, const Mod b) { return a = a / b; }
Mod fact[1024000], factinv[1024000];
void init(const int amax = 1024000) {
  fact[0] = Mod(1);
  factinv[0] = 1;
  for (int i = 0; i < amax - 1; ++i) {
    fact[i + 1] = fact[i] * Mod(i + 1);
    factinv[i + 1] = factinv[i] / Mod(i + 1);
  }
}
Mod comb(const int a, const int b) {
  return fact[a] * factinv[b] * factinv[a - b];
}
using ll = long long int;
Mod solve(ll x, int kk) {
  string st = to_string(x);
  vector<vector<vector<pair<Mod, Mod>>>> memo(
      st.size() + 1,
      vector<vector<pair<Mod, Mod>>>(
          1 << 10, vector<pair<Mod, Mod>>(2, make_pair(Mod(0), Mod(0)))));
  memo[st.size()][0][1] = make_pair(Mod(1), Mod(0));
  for (int now = st.size() - 1; now >= 0; --now) {
    for (int use = 0; use < 1024; ++use) {
      for (int same = 0; same < 2; ++same) {
        for (int now_use = 0; now_use < 10; ++now_use) {
          if (same && now_use > st[st.size() - 1 - now] - '0')
            continue;
          else {
            int next_use = use | (1 << now_use);
            if (now_use == 0 && use == 0) {
              next_use = 0;
            }
            int next_same = same && (now_use == st[st.size() - 1 - now] - '0');
            memo[now][next_use][next_same].first +=
                memo[now + 1][use][same].first;
            memo[now][next_use][next_same].second +=
                memo[now + 1][use][same].second +
                memo[now + 1][use][same].first * now_use * mod_pow(10, now);
          }
        }
      }
    }
  }
  Mod answer = 0;
  for (int i = 0; i < 1024; ++i) {
    for (int same = 0; same < 2; ++same) {
      if (memo[0][i][same].second.num > 0) {
      }
      if (__builtin_popcount(i) <= kk) answer += memo[0][i][same].second;
    }
  }
  return answer;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  ll l, r, k;
  cin >> l >> r >> k;
  Mod ans1 = solve(r, k);
  Mod ans2 = solve(l - 1, k);
  cout << (ans1 - ans2).num << endl;
  return 0;
}
