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;
inline long long quickpow(long long m, long long n, long long p) {
  long long b = 1;
  while (n) {
    if (n & 1) b = b * m % p;
    n = n >> 1;
    m = m * m % p;
  }
  return b;
}
inline long long quickmul(long long x, long long y, long long mod) {
  return (x * y - (long long)((long double)x / mod * y) * mod + mod) % mod;
}
inline long long getinv(long long x, long long p) {
  return quickpow(x, p - 2, p);
}
inline long long read(void) {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    f = ch == '-' ? -1 : 1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long a[21], len, k;
pair<long long, long long> f[21][(1 << 11)];
long long vis[21][(1 << 11)], pw[21];
const long long Mod = 998244353;
inline pair<long long, long long> dfs(long long pos, long long stat,
                                      long long pre, long long lim) {
  if (pos == 0) {
    long long rret = 0;
    for (long long j = 0; j <= 9; ++j) rret += (stat >> j) & 1;
    return rret > k ? make_pair(0, 0) : make_pair(1, 0);
  }
  if (!lim && !pre && vis[pos][stat] != -1) return f[pos][stat];
  long long up = lim ? a[pos] : 9;
  pair<long long, long long> ans = make_pair(0, 0);
  for (long long i = 0; i <= up; ++i) {
    long long sta = (i == 0 && pre) ? 0 : (1 << i) | stat;
    pair<long long, long long> tmp =
        dfs(pos - 1, sta, i == 0 && pre, lim && i == a[pos]);
    long long p = i * pw[pos - 1] % Mod;
    ans.first = (ans.first + tmp.first) % Mod;
    ans.second = (ans.second + tmp.second + tmp.first * p % Mod) % Mod;
  }
  if (!lim && !pre) f[pos][stat] = ans, vis[pos][stat] = 1;
  return ans;
}
inline long long solve(long long x) {
  len = 0;
  while (x) a[++len] = x % 10, x /= 10;
  long long d = dfs(len, 0, 1, 1).second;
  return d;
}
signed main(signed argc, char *argv[]) {
  memset(vis, -1, sizeof(vis));
  pw[0] = 1;
  for (long long i = 1; i <= 20; i++) pw[i] = pw[i - 1] * 10 % Mod;
  long long l = read(), r = read();
  k = read();
  printf("%lld\n", (solve(r) - solve(l - 1) + Mod) % Mod);
  return 0;
}
