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 M = 998244353;
const int N = 20;
struct re {
  int x, y;
};
long long L, R;
int K, na, nb;
int a[N], b[N];
void init(long long X, int *a, int &n) {
  while (X) {
    a[n++] = X % 10;
    X /= 10;
  }
}
long long f[N][1 << 10], g[N][1 << 10], pw[N];
void initfg() {
  pw[0] = 1;
  for (int i = (1); (i <= (18)); ++i) pw[i] = pw[i - 1] * 10 % M;
  f[0][0] = 1;
  for (int i = (0); (i <= (17)); ++i) {
    for (int mask = (0); (mask <= ((1 << 10) - 1)); ++mask)
      if (f[i][mask]) {
        for (int j = (0); (j <= (9)); ++j) {
          f[i + 1][mask | (1 << j)] += f[i][mask];
          f[i + 1][mask | (1 << j)] %= M;
          g[i + 1][mask | (1 << j)] += g[i][mask] + f[i][mask] * j * pw[i] % M;
          g[i + 1][mask | (1 << j)] %= M;
        }
      }
  }
}
long long solve(int *a, int n) {
  long long res = 0, cursum = 0;
  int curmask = 0;
  bool ok = 0;
  while (n--) {
    if (n) {
      for (int j = (1); (j <= (9)); ++j)
        for (int mask = (0); (mask <= ((1 << 10) - 1)); ++mask)
          if (__builtin_popcount((1 << j) | mask) <= K && f[n - 1][mask]) {
            res += g[n - 1][mask] + f[n - 1][mask] * j * pw[n - 1] % M;
            res %= M;
          }
    }
    for (int j = (1 - ok); (j <= (a[n] - 1)); ++j)
      for (int mask = (0); (mask <= ((1 << 10) - 1)); ++mask)
        if (__builtin_popcount(curmask | (1 << j) | mask) <= K && f[n][mask]) {
          res = res + g[n][mask] % M +
                f[n][mask] * ((j * pw[n] + cursum) % M) % M;
          res %= M;
        }
    curmask |= (1 << a[n]);
    cursum += pw[n] * a[n] % M;
    cursum %= M;
    ok = 1;
  }
  if (__builtin_popcount(curmask) <= K) res += cursum, res %= M;
  return res;
}
int main() {
  cin >> L >> R >> K;
  init(R, a, na);
  init(L - 1, b, nb);
  initfg();
  cout << (solve(a, na) - solve(b, nb) + M) % M;
}
