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;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using ll = long long;
using num = modnum<998244353>;
using arr = array<pair<num, num>, 1024>;
int mask(ll v) {
  int r = 0;
  while (v) {
    r |= 1 << (v % 10);
    v /= 10;
  }
  return r;
}
arr dfs(ll a, ll b) {
  arr res = {};
  assert(a < b);
  if (a / 10 < b / 10) {
    arr tmp = dfs(a / 10, b / 10);
    for (int m = 0; m < 1024; m++) {
      for (int d = 0; d < 10; d++) {
        int nm = m | (1 << d);
        res[nm].first += tmp[m].first;
        res[nm].second += 10 * tmp[m].second + tmp[m].first * d;
      }
    }
    for (ll v = a + 1; v < a / 10 * 10 + 10; v++) {
      int m = mask(v);
      res[m].first += 1;
      res[m].second += v;
    }
    for (ll v = b / 10 * 10; v < b; v++) {
      int m = mask(v);
      res[m].first += 1;
      res[m].second += v;
    }
  } else {
    for (ll v = a + 1; v < b; v++) {
      int m = mask(v);
      res[m].first += 1;
      res[m].second += v;
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  ll L, R;
  cin >> L >> R;
  L--, R++;
  int K;
  cin >> K;
  arr res = dfs(L, R);
  num ans = 0;
  for (int m = 0; m < 1024; m++) {
    if (__builtin_popcount(m) <= K) ans += res[m].second;
  }
  cout << ans << '\n';
  return 0;
}
