#include <bits/stdc++.h>
using namespace std;
const int N = 3e5, M = 998244353;
long long l, r;
int k, freq[1 << 10], p[20];
vector<char> v;
pair<int, int> dp[18][2][1 << 11];
void intial() {
  for (int i = 0; i < 18; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < (1 << 10); k++) dp[i][j][k] = {-1, -1};
}
void add(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
  if (a < 0) a += M;
}
pair<int, int> calc(int u, bool is, int did) {
  if (freq[did] > k) return make_pair(0, 0);
  if (u < 0) return make_pair(0, 1);
  pair<int, int> &ref = dp[u][is][did];
  if (ref != make_pair(-1, -1)) return ref;
  ref = {0, 0};
  int to = !is ? v[u] : 9;
  for (int i = 0; i <= to; i++) {
    int lol = did;
    if (i || !i && did) lol |= 1 << i;
    pair<int, int> temp = calc(u - 1, is | (i < v[u]), lol);
    add(ref.first, (i * 1LL * p[u] % M) * temp.second % M);
    add(ref.first, temp.first);
    add(ref.second, temp.second);
  }
  return ref;
}
int main() {
  p[0] = 1;
  for (int i = 1; i < 20; i++) p[i] = p[i - 1] * 10LL % M;
  for (int i = 0; i < (1 << 10); i++) freq[i] = __builtin_popcount(i);
  scanf("%lld%lld%d", &l, &r, &k);
  intial();
  while (v.size() < 18) v.push_back(r % 10), r /= 10;
  int ans = calc(17, 0, 0).first;
  v.clear();
  intial();
  long long temp = l;
  vector<int> has;
  while (v.size() < 18) {
    if (l) has.push_back(l % 10);
    v.push_back(l % 10), l /= 10;
  }
  sort(has.begin(), has.end());
  has.resize(unique(has.begin(), has.end()) - has.begin());
  if (has.size() <= k) ans = (ans + temp) % M;
  printf("%d\n", ((ans - calc(17, 0, 0).first) % M + M) % M);
}

### Prompt

Your challenge is to write 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 N = 3e5, M = 998244353;
long long l, r;
int k, freq[1 << 10], p[20];
vector<char> v;
pair<int, int> dp[18][2][1 << 11];
void intial() {
  for (int i = 0; i < 18; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < (1 << 10); k++) dp[i][j][k] = {-1, -1};
}
void add(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
  if (a < 0) a += M;
}
pair<int, int> calc(int u, bool is, int did) {
  if (freq[did] > k) return make_pair(0, 0);
  if (u < 0) return make_pair(0, 1);
  pair<int, int> &ref = dp[u][is][did];
  if (ref != make_pair(-1, -1)) return ref;
  ref = {0, 0};
  int to = !is ? v[u] : 9;
  for (int i = 0; i <= to; i++) {
    int lol = did;
    if (i || !i && did) lol |= 1 << i;
    pair<int, int> temp = calc(u - 1, is | (i < v[u]), lol);
    add(ref.first, (i * 1LL * p[u] % M) * temp.second % M);
    add(ref.first, temp.first);
    add(ref.second, temp.second);
  }
  return ref;
}
int main() {
  p[0] = 1;
  for (int i = 1; i < 20; i++) p[i] = p[i - 1] * 10LL % M;
  for (int i = 0; i < (1 << 10); i++) freq[i] = __builtin_popcount(i);
  scanf("%lld%lld%d", &l, &r, &k);
  intial();
  while (v.size() < 18) v.push_back(r % 10), r /= 10;
  int ans = calc(17, 0, 0).first;
  v.clear();
  intial();
  long long temp = l;
  vector<int> has;
  while (v.size() < 18) {
    if (l) has.push_back(l % 10);
    v.push_back(l % 10), l /= 10;
  }
  sort(has.begin(), has.end());
  has.resize(unique(has.begin(), has.end()) - has.begin());
  if (has.size() <= k) ans = (ans + temp) % M;
  printf("%d\n", ((ans - calc(17, 0, 0).first) % M + M) % M);
}

```