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 N = 20, MOD = 998244353;
long long n, m, k;
long long a[N], dp[N][1 << 12], f[N][1 << 12], inv[N];
struct node {
  long long num, sum;
};
node dfs(long long x, long long p, long long cnt, long long lead,
         long long limit) {
  if (x == 0) {
    if (lead)
      return (node){0, 0};
    else
      return (node){1, 0};
  }
  if (!lead && !limit && dp[x][p] != -1) return (node){dp[x][p], f[x][p]};
  long long h;
  if (limit)
    h = a[x];
  else
    h = 9;
  node res, now;
  res.num = res.sum = now.num = now.sum = 0;
  for (register long long i = 0; i <= h; ++i) {
    if (cnt == k && ((1ll << i) & p) == 0) continue;
    if (lead && (i == 0)) {
      now = dfs(x - 1, p, cnt, 1, (i == h) && limit);
      res.num = (res.num + now.num) % MOD;
      res.sum = (res.sum + now.sum) % MOD;
      res.sum = (res.sum + i * inv[x] % MOD * now.num % MOD) % MOD;
      continue;
    }
    if ((1ll << i) & p) {
      now = dfs(x - 1, p, cnt, 0, (i == h) && limit);
      res.num = (res.num + now.num) % MOD;
      res.sum = (res.sum + now.sum) % MOD;
      res.sum = (res.sum + i * inv[x] % MOD * now.num % MOD) % MOD;
    } else {
      now = dfs(x - 1, (1ll << i) | p, cnt + 1, 0, (i == h) && limit);
      res.num = (res.num + now.num) % MOD;
      res.sum = (res.sum + now.sum) % MOD;
      res.sum = (res.sum + i * inv[x] % MOD * now.num % MOD) % MOD;
    }
  }
  if (!limit && !lead) {
    dp[x][p] = res.num;
    f[x][p] = res.sum;
  }
  return res;
}
inline long long solve(long long x) {
  memset(dp, -1, sizeof(dp));
  memset(f, 0, sizeof(f));
  long long len = 0;
  while (x) {
    a[++len] = x % 10;
    x /= 10;
  }
  return dfs(len, 0, 0, 1, 1).sum;
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  inv[1] = 1;
  for (register long long i = 2; i <= 20; ++i) inv[i] = inv[i - 1] * 10 % MOD;
  printf("%lld\n", ((solve(m) - solve(n - 1)) % MOD + MOD) % MOD);
  return 0;
}
