#include <bits/stdc++.h>
const int mod = 998244353;
int tot, num[20];
long long k, l, r;
inline void up(long long &a, long long b) {
  if ((a += b) >= mod) a -= mod;
}
struct node {
  long long cnt, sum;
} f[20][1 << 10];
long long pw[20];
node calc(int x, int lim, int S) {
  if (!x) return (node){1, 0};
  if (!lim && ~f[x][S].cnt) return f[x][S];
  node F = (node){0, 0};
  for (int i = lim ? num[x] : 9; ~i; i--) {
    int nxt;
    if (!S && !i)
      nxt = 0;
    else
      nxt = S | 1 << i;
    if (__builtin_popcount(nxt) > k) continue;
    node tmp = calc(x - 1, lim && i == num[x], nxt);
    up(F.cnt, tmp.cnt);
    up(F.sum, (tmp.sum + tmp.cnt * pw[x - 1] % mod * i) % mod);
  }
  if (!lim) f[x][S] = F;
  return F;
}
long long solve(long long x) {
  tot = 0;
  while (x) {
    num[++tot] = x % 10;
    x /= 10;
  }
  return calc(tot, 1, 0).sum;
}
int main() {
  __builtin_memset(f, -1, sizeof f);
  scanf("%lld%lld%lld", &l, &r, &k);
  pw[0] = 1;
  for (int i = 1; i < 20; i++) pw[i] = pw[i - 1] * 10 % mod;
  printf("%lld\n", (solve(r) - solve(l - 1) + mod) % mod);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
const int mod = 998244353;
int tot, num[20];
long long k, l, r;
inline void up(long long &a, long long b) {
  if ((a += b) >= mod) a -= mod;
}
struct node {
  long long cnt, sum;
} f[20][1 << 10];
long long pw[20];
node calc(int x, int lim, int S) {
  if (!x) return (node){1, 0};
  if (!lim && ~f[x][S].cnt) return f[x][S];
  node F = (node){0, 0};
  for (int i = lim ? num[x] : 9; ~i; i--) {
    int nxt;
    if (!S && !i)
      nxt = 0;
    else
      nxt = S | 1 << i;
    if (__builtin_popcount(nxt) > k) continue;
    node tmp = calc(x - 1, lim && i == num[x], nxt);
    up(F.cnt, tmp.cnt);
    up(F.sum, (tmp.sum + tmp.cnt * pw[x - 1] % mod * i) % mod);
  }
  if (!lim) f[x][S] = F;
  return F;
}
long long solve(long long x) {
  tot = 0;
  while (x) {
    num[++tot] = x % 10;
    x /= 10;
  }
  return calc(tot, 1, 0).sum;
}
int main() {
  __builtin_memset(f, -1, sizeof f);
  scanf("%lld%lld%lld", &l, &r, &k);
  pw[0] = 1;
  for (int i = 1; i < 20; i++) pw[i] = pw[i - 1] * 10 % mod;
  printf("%lld\n", (solve(r) - solve(l - 1) + mod) % mod);
  return 0;
}

```