#include <bits/stdc++.h>
using namespace std;
const long long modd = 998244353;
int num[22];
int k;
struct Node {
  long long sum, num;
} dp[22][(1 << 10) + 10];
bool vis[22][(1 << 10) + 10];
long long pw10[22];
int cal(long long x) {
  int res = 0;
  while (x) {
    if (x & 1) ++res;
    x >>= 1;
  }
  return res;
}
Node dfs(int pos, int s, bool fg, bool all0) {
  if (pos == 0) {
    Node tem;
    tem.sum = 0;
    tem.num = 1;
    return tem;
  }
  if (fg == 0 && all0 == 0 && vis[pos][s]) return dp[pos][s];
  int lim = fg ? num[pos] : 9;
  Node ans;
  ans.sum = 0;
  ans.num = 0;
  for (int i = 0; i <= lim; ++i) {
    if (all0 && i == 0) {
      ans.sum =
          (ans.sum + dfs(pos - 1, s, fg && i == lim, all0 && i == 0).sum) %
          modd;
      ans.num =
          (ans.num + dfs(pos - 1, s, fg && i == lim, all0 && i == 0).num) %
          modd;
    } else {
      int cnt = cal(s | (1 << i));
      if (cnt > k) continue;
      Node tem = dfs(pos - 1, s | (1 << i), fg && i == lim, all0 && i == 0);
      ans.sum =
          (ans.sum + tem.sum + pw10[pos - 1] * i % modd * tem.num % modd) %
          modd;
      ans.num = (ans.num + tem.num) % modd;
    }
  }
  if (fg == 0 && all0 == 0) {
    vis[pos][s] = 1;
    return dp[pos][s] = ans;
  }
  return ans;
}
long long solve(long long x) {
  for (int i = 0; i <= 20; ++i) num[i] = 0;
  int cnt = 1;
  for (; x; ++cnt) {
    num[cnt] = x % 10;
    x /= 10;
  }
  memset(vis, 0, sizeof(vis));
  return dfs(cnt, 0, 1, 1).sum;
}
int main() {
  pw10[0] = 1;
  for (int i = 1; i <= 19; ++i) pw10[i] = pw10[i - 1] * 10 % modd;
  long long L, R;
  scanf("%lld %lld %d", &L, &R, &k);
  printf("%lld", (solve(R) - solve(L - 1) + modd) % modd);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long modd = 998244353;
int num[22];
int k;
struct Node {
  long long sum, num;
} dp[22][(1 << 10) + 10];
bool vis[22][(1 << 10) + 10];
long long pw10[22];
int cal(long long x) {
  int res = 0;
  while (x) {
    if (x & 1) ++res;
    x >>= 1;
  }
  return res;
}
Node dfs(int pos, int s, bool fg, bool all0) {
  if (pos == 0) {
    Node tem;
    tem.sum = 0;
    tem.num = 1;
    return tem;
  }
  if (fg == 0 && all0 == 0 && vis[pos][s]) return dp[pos][s];
  int lim = fg ? num[pos] : 9;
  Node ans;
  ans.sum = 0;
  ans.num = 0;
  for (int i = 0; i <= lim; ++i) {
    if (all0 && i == 0) {
      ans.sum =
          (ans.sum + dfs(pos - 1, s, fg && i == lim, all0 && i == 0).sum) %
          modd;
      ans.num =
          (ans.num + dfs(pos - 1, s, fg && i == lim, all0 && i == 0).num) %
          modd;
    } else {
      int cnt = cal(s | (1 << i));
      if (cnt > k) continue;
      Node tem = dfs(pos - 1, s | (1 << i), fg && i == lim, all0 && i == 0);
      ans.sum =
          (ans.sum + tem.sum + pw10[pos - 1] * i % modd * tem.num % modd) %
          modd;
      ans.num = (ans.num + tem.num) % modd;
    }
  }
  if (fg == 0 && all0 == 0) {
    vis[pos][s] = 1;
    return dp[pos][s] = ans;
  }
  return ans;
}
long long solve(long long x) {
  for (int i = 0; i <= 20; ++i) num[i] = 0;
  int cnt = 1;
  for (; x; ++cnt) {
    num[cnt] = x % 10;
    x /= 10;
  }
  memset(vis, 0, sizeof(vis));
  return dfs(cnt, 0, 1, 1).sum;
}
int main() {
  pw10[0] = 1;
  for (int i = 1; i <= 19; ++i) pw10[i] = pw10[i - 1] * 10 % modd;
  long long L, R;
  scanf("%lld %lld %d", &L, &R, &k);
  printf("%lld", (solve(R) - solve(L - 1) + modd) % modd);
  return 0;
}

```