#include <bits/stdc++.h>
using namespace std;
const int MAXN = 18 + 5;
const int ALL = (1 << 10) + 5;
const int all = (1 << 10) - 1;
const int mod = 998244353;
inline void add_mod(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int pw10[MAXN];
int tot[ALL];
int n, d;
char s[MAXN];
pair<int, int> f[MAXN][ALL];
bool vis[MAXN][ALL];
pair<int, int> gao(int pos, int mask, bool lead, bool lim) {
  if (pos > n) return make_pair(1, 0);
  if (vis[pos][mask] && !lead && !lim) return f[pos][mask];
  pair<int, int> res = make_pair(0, 0);
  int up = lim ? s[pos] : 9;
  for (int i = 0; i <= up; ++i) {
    int now = (lead && !i) ? 0 : (mask | (1 << i));
    if (tot[now] > d) continue;
    pair<int, int> tmp = gao(pos + 1, now, lead && !i, lim && i == up);
    add_mod(res.first, tmp.first);
    add_mod(res.second,
            ((long long)tmp.first * i * pw10[n - pos] + tmp.second) % mod);
  }
  if (!lim && !lead) f[pos][mask] = res, vis[pos][mask] = 1;
  return res;
}
int calc(long long x) {
  memset(vis, 0, sizeof(vis));
  sprintf(s + 1, "%lld", x);
  n = strlen(s + 1);
  for (int i = 1; i <= n; ++i) s[i] -= '0';
  return gao(1, 0, 1, 1).second;
}
int main(void) {
  pw10[0] = 1;
  for (int i = 1; i < MAXN; ++i) pw10[i] = (long long)pw10[i - 1] * 10 % mod;
  tot[0] = 0;
  for (int i = 1; i < ALL; ++i) tot[i] = tot[i ^ ((i) & -(i))] + 1;
  long long l, r;
  scanf("%lld%lld%d", &l, &r, &d);
  int ans = calc(r) - calc(l - 1);
  printf("%d", (ans % mod + mod) % mod);
  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 int MAXN = 18 + 5;
const int ALL = (1 << 10) + 5;
const int all = (1 << 10) - 1;
const int mod = 998244353;
inline void add_mod(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int pw10[MAXN];
int tot[ALL];
int n, d;
char s[MAXN];
pair<int, int> f[MAXN][ALL];
bool vis[MAXN][ALL];
pair<int, int> gao(int pos, int mask, bool lead, bool lim) {
  if (pos > n) return make_pair(1, 0);
  if (vis[pos][mask] && !lead && !lim) return f[pos][mask];
  pair<int, int> res = make_pair(0, 0);
  int up = lim ? s[pos] : 9;
  for (int i = 0; i <= up; ++i) {
    int now = (lead && !i) ? 0 : (mask | (1 << i));
    if (tot[now] > d) continue;
    pair<int, int> tmp = gao(pos + 1, now, lead && !i, lim && i == up);
    add_mod(res.first, tmp.first);
    add_mod(res.second,
            ((long long)tmp.first * i * pw10[n - pos] + tmp.second) % mod);
  }
  if (!lim && !lead) f[pos][mask] = res, vis[pos][mask] = 1;
  return res;
}
int calc(long long x) {
  memset(vis, 0, sizeof(vis));
  sprintf(s + 1, "%lld", x);
  n = strlen(s + 1);
  for (int i = 1; i <= n; ++i) s[i] -= '0';
  return gao(1, 0, 1, 1).second;
}
int main(void) {
  pw10[0] = 1;
  for (int i = 1; i < MAXN; ++i) pw10[i] = (long long)pw10[i - 1] * 10 % mod;
  tot[0] = 0;
  for (int i = 1; i < ALL; ++i) tot[i] = tot[i ^ ((i) & -(i))] + 1;
  long long l, r;
  scanf("%lld%lld%d", &l, &r, &d);
  int ans = calc(r) - calc(l - 1);
  printf("%d", (ans % mod + mod) % mod);
  return 0;
}

```