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 mod = 998244353;
long long dp[20][1024][12], sum[20][1024][12];
int a[20];
long long pw10[19];
long long dfsnum(int pos, int zero, int sta, int limit, int rnum) {
  if (rnum < 0) return 0;
  if (pos == -1) return 1;
  if (!limit && dp[pos][sta][rnum] != -1) return dp[pos][sta][rnum];
  int up = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    if (i == 0 && !zero)
      ans += dfsnum(pos - 1, zero, sta, limit && i == a[pos], rnum);
    else if (((1 << i) & sta) == 0)
      ans += dfsnum(pos - 1, zero | 1, sta | (1 << i), limit && i == a[pos],
                    rnum - 1);
    else
      ans += dfsnum(pos - 1, zero | 1, sta, limit && i == a[pos], rnum);
    ans %= mod;
  }
  ans %= mod;
  if (!limit) dp[pos][sta][rnum] = ans;
  return ans;
}
long long dfssum(int pos, int zero, int sta, int limit, int rnum) {
  if (rnum < 0) return 0;
  if (pos == -1) return 0;
  if (!limit && sum[pos][sta][rnum] != -1) return sum[pos][sta][rnum];
  int up = limit ? a[pos] : 9;
  long long ans = 0, tmp;
  for (int i = 0; i <= up; i++) {
    if (i == 0 && !zero)
      ans += dfssum(pos - 1, zero, sta, limit && i == a[pos], rnum);
    else if (((1 << i) & sta) == 0) {
      tmp = dfsnum(pos - 1, zero | 1, sta | (1 << i), limit && i == a[pos],
                   rnum - 1);
      ans +=
          tmp * i * pw10[pos] % mod + dfssum(pos - 1, zero | 1, sta | (1 << i),
                                             limit && i == a[pos], rnum - 1);
    } else {
      tmp = dfsnum(pos - 1, zero | 1, sta, limit && i == a[pos], rnum);
      ans += tmp * i * pw10[pos] % mod +
             dfssum(pos - 1, zero | 1, sta, limit && i == a[pos], rnum);
    }
    ans %= mod;
  }
  ans %= mod;
  if (!limit) sum[pos][sta][rnum] = ans;
  return ans;
}
long long solve(long long x, int rnum) {
  int pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfssum(pos - 1, 0, 0, 1, rnum);
}
int main() {
  long long tt, n, m, i, j, k, s, t, l, r;
  int rnum;
  memset(dp, 255, sizeof(dp));
  memset(sum, 255, sizeof(sum));
  pw10[0] = 1;
  for (i = 1; i <= 18; i++) pw10[i] = (pw10[i - 1] * 10) % mod;
  scanf("%I64d%I64d%d", &l, &r, &rnum);
  printf("%I64d\n", (mod + solve(r, rnum) - solve(l - 1, rnum)) % mod);
  return 0;
}
