#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
struct P {
  long long cnt, sum;
} dp[19][1025];
long long k, l, r;
long long a[19];
long long get_st(long long st) {
  long long sum = 0;
  for (int i = 0; i <= 9; i++)
    if (st & (1 << i)) sum++;
  return sum;
}
long long _pow(long long ci) {
  long long ans = 1, x = 10;
  while (ci) {
    if (ci & 1) ans = (ans * x) % mod;
    ci >>= 1;
    x = (x * x) % mod;
  }
  return ans % mod;
}
P dfs(long long now, long long st, bool limit, bool lead) {
  if (!now) return P{get_st(st) <= k, 0};
  if (!lead && !limit && dp[now][st].cnt != -1) return dp[now][st];
  long long up = limit ? a[now] : 9;
  P tot = {0, 0};
  for (int i = 0; i <= up; i++) {
    long long new_st = (lead && i == 0) ? st : st | (1 << i);
    P tmp = dfs(now - 1, new_st, limit && i == a[now], lead && i == 0);
    tot = P{
        (tot.cnt + tmp.cnt) % mod,
        ((tot.sum + tmp.sum) % mod + _pow(now - 1) * i % mod * tmp.cnt % mod) %
            mod};
  }
  if (!lead && !limit) dp[now][st] = tot;
  return tot;
}
long long solve(long long num) {
  long long pos = 0, ans = 0;
  while (num) {
    a[++pos] = num % 10;
    num /= 10;
  }
  return dfs(pos, 0, 1, 1).sum % mod;
}
int main() {
  for (int i = 1; i <= 18; i++)
    for (int j = 1; j <= 1024; j++) dp[i][j].cnt = -1;
  scanf("%lld%lld%lld", &l, &r, &k);
  printf("%lld", (solve(r) - solve(l - 1) + mod) % mod);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 mod = 998244353;
struct P {
  long long cnt, sum;
} dp[19][1025];
long long k, l, r;
long long a[19];
long long get_st(long long st) {
  long long sum = 0;
  for (int i = 0; i <= 9; i++)
    if (st & (1 << i)) sum++;
  return sum;
}
long long _pow(long long ci) {
  long long ans = 1, x = 10;
  while (ci) {
    if (ci & 1) ans = (ans * x) % mod;
    ci >>= 1;
    x = (x * x) % mod;
  }
  return ans % mod;
}
P dfs(long long now, long long st, bool limit, bool lead) {
  if (!now) return P{get_st(st) <= k, 0};
  if (!lead && !limit && dp[now][st].cnt != -1) return dp[now][st];
  long long up = limit ? a[now] : 9;
  P tot = {0, 0};
  for (int i = 0; i <= up; i++) {
    long long new_st = (lead && i == 0) ? st : st | (1 << i);
    P tmp = dfs(now - 1, new_st, limit && i == a[now], lead && i == 0);
    tot = P{
        (tot.cnt + tmp.cnt) % mod,
        ((tot.sum + tmp.sum) % mod + _pow(now - 1) * i % mod * tmp.cnt % mod) %
            mod};
  }
  if (!lead && !limit) dp[now][st] = tot;
  return tot;
}
long long solve(long long num) {
  long long pos = 0, ans = 0;
  while (num) {
    a[++pos] = num % 10;
    num /= 10;
  }
  return dfs(pos, 0, 1, 1).sum % mod;
}
int main() {
  for (int i = 1; i <= 18; i++)
    for (int j = 1; j <= 1024; j++) dp[i][j].cnt = -1;
  scanf("%lld%lld%lld", &l, &r, &k);
  printf("%lld", (solve(r) - solve(l - 1) + mod) % mod);
}

```