#include <bits/stdc++.h>
using namespace std;
inline long long Min(long long x, long long y) { return x < y ? x : y; }
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
bool cmp(long long x, long long y) { return x > y; }
long long k, a[20], Jw[20];
struct node {
  long long tot, Sum;
  node() {}
  node(long long TOT, long long SUM) {
    tot = TOT;
    Sum = SUM;
  }
  void Clear() { tot = Sum = 0; }
  bool Check() { return ~tot && ~Sum; }
} dp[2][2][20][(1 << 10)];
void Add(node &a, node b, long long Num, long long Pos) {
  a.tot = (a.tot + b.tot) % 998244353;
  a.Sum = (a.Sum + Num * Jw[Pos] % 998244353 * b.tot % 998244353 + b.Sum) %
          998244353;
}
node Dfs(bool Limited, bool Zero, long long Pos, long long Zt) {
  if (!Pos) return node(1, 0);
  node &Res = dp[Limited][Zero][Pos][Zt];
  if (Res.Check()) return Res;
  Res.Clear();
  for (register int i = 0; i <= (Limited ? a[Pos] : 9); ++i)
    if (__builtin_popcount(Zero && !i ? 0 : Zt | 1 << i) <= k)
      Add(Res,
          Dfs(Limited && i == a[Pos], Zero && !i, Pos - 1,
              Zero && !i ? 0 : Zt | 1 << i),
          i, Pos - 1);
  return Res;
}
long long Solve(long long x) {
  long long Len = 0;
  memset(dp, -1, sizeof dp);
  while (x) a[++Len] = x % 10, x /= 10;
  return Dfs(1, 1, Len, 0).Sum;
}
int main() {
  Jw[0] = 1;
  for (register int i = 1; i < 19; ++i) Jw[i] = Jw[i - 1] * 10 % 998244353;
  long long l, r;
  read(l);
  read(r);
  read(k);
  printf("%lld", (Solve(r) - Solve(l - 1) + 998244353) % 998244353);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
inline long long Min(long long x, long long y) { return x < y ? x : y; }
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
bool cmp(long long x, long long y) { return x > y; }
long long k, a[20], Jw[20];
struct node {
  long long tot, Sum;
  node() {}
  node(long long TOT, long long SUM) {
    tot = TOT;
    Sum = SUM;
  }
  void Clear() { tot = Sum = 0; }
  bool Check() { return ~tot && ~Sum; }
} dp[2][2][20][(1 << 10)];
void Add(node &a, node b, long long Num, long long Pos) {
  a.tot = (a.tot + b.tot) % 998244353;
  a.Sum = (a.Sum + Num * Jw[Pos] % 998244353 * b.tot % 998244353 + b.Sum) %
          998244353;
}
node Dfs(bool Limited, bool Zero, long long Pos, long long Zt) {
  if (!Pos) return node(1, 0);
  node &Res = dp[Limited][Zero][Pos][Zt];
  if (Res.Check()) return Res;
  Res.Clear();
  for (register int i = 0; i <= (Limited ? a[Pos] : 9); ++i)
    if (__builtin_popcount(Zero && !i ? 0 : Zt | 1 << i) <= k)
      Add(Res,
          Dfs(Limited && i == a[Pos], Zero && !i, Pos - 1,
              Zero && !i ? 0 : Zt | 1 << i),
          i, Pos - 1);
  return Res;
}
long long Solve(long long x) {
  long long Len = 0;
  memset(dp, -1, sizeof dp);
  while (x) a[++Len] = x % 10, x /= 10;
  return Dfs(1, 1, Len, 0).Sum;
}
int main() {
  Jw[0] = 1;
  for (register int i = 1; i < 19; ++i) Jw[i] = Jw[i - 1] * 10 % 998244353;
  long long l, r;
  read(l);
  read(r);
  read(k);
  printf("%lld", (Solve(r) - Solve(l - 1) + 998244353) % 998244353);
  return 0;
}

```