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;
int cnt[20];
long long d[20];
long long l, r, k;
struct Point {
  long long x, y;
} dp[20][1 << 11][2];
Point dfs(long long len, long long state, bool limit, bool non_zero) {
  if (len == 0) return Point{1, 0};
  if (!limit && dp[len][state][non_zero].y) return dp[len][state][non_zero];
  Point ans = Point{0, 0};
  int maxn = limit ? cnt[len] : 9;
  for (int i = 0; i <= maxn; ++i) {
    long long temp = state | ((non_zero || i) << i);
    if (__builtin_popcountll(temp) > k) continue;
    Point t = dfs(len - 1, temp, limit && i == maxn, non_zero || i);
    ans.x = (ans.x + t.x) % MOD;
    ans.y = (ans.y + t.y + i * d[len - 1] % MOD * t.x % MOD) % MOD;
  }
  return dp[len][state][non_zero] = ans;
}
long long solve(long long x) {
  memset(dp, 0, sizeof dp);
  memset(cnt, 0, sizeof cnt);
  int len = 0;
  while (x) {
    cnt[++len] = x % 10;
    x /= 10;
  }
  return dfs(len, 0, true, 0).y;
}
int main() {
  d[0] = 1;
  for (int i = 1; i < 20; ++i) d[i] = d[i - 1] * 10 % MOD;
  cin >> l >> r >> k;
  cout << (solve(r) - solve(l - 1) + MOD) % MOD << endl;
  return 0;
}
