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 int inf = 0x3c3c3c3c;
const long long infl = 0x3c3c3c3c3c3c3c3c;
const long long mod = 998244353;
long long L, R, K;
pair<long long, long long> dp[18][1 << 10][2];
long long X;
int num[20];
long long md[20];
pair<long long, long long> go(int pos, int mask, int shit) {
  if (pos == -1) return {1, 0};
  pair<long long, long long>& cache = dp[pos][mask][shit];
  if (cache.first != -1) return cache;
  long long ret = 0;
  int cnt = 0;
  for (int i = 0; i < 10; i++) cnt += ((mask >> i) & 1);
  long long ret1 = 0, ret2 = 0;
  for (int i = 0; i < 10; i++) {
    if (!shit && num[pos] < i) continue;
    int add = ((~(mask >> i)) & 1);
    if (cnt + add > K) continue;
    int n_mask = (mask | (1 << i));
    if (mask == 0 && i == 0) n_mask = 0;
    int n_shit = (shit || (i < num[pos]));
    auto g = go(pos - 1, n_mask, n_shit);
    long long tmp = (md[pos] * i) % mod * g.first % mod;
    ret1 = (ret1 + g.first) % mod;
    ret2 = ((ret2 + tmp) % mod + g.second) % mod;
  }
  return cache = {ret1, ret2};
}
void init(long long x) {
  for (int i = 0; i < 18; i++) {
    for (int j = 0; j < (1 << 10); j++) {
      for (int k = 0; k < 2; k++) {
        dp[i][j][k] = {-1, -1};
      }
    }
  }
  memset(num, 0, sizeof(num));
  X = x;
  int idx = 0;
  while (x) {
    num[idx] = x % 10;
    x /= 10;
    idx++;
  }
}
int main() {
  cin >> L >> R >> K;
  md[0] = 1;
  for (int i = 1; i < 20; i++) {
    md[i] = (md[i - 1] * 10) % mod;
  }
  init(L - 1);
  long long lo = go(17, 0, 0).second;
  init(R);
  long long hi = go(17, 0, 0).second;
  cout << (hi - lo + mod) % mod << endl;
  return 0;
}
