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.
#include <bits/stdc++.h>
const int MOD = 998244353;
long long l, r;
int k;
int bit[25];
int getb[1 << 10];
int powt[25];
struct node {
  bool vis;
  int cnt, sum;
} dp[25][1 << 10];
void inc(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
node dfs(int pos, int mask, bool zero, bool limit) {
  if (pos == -1) return (node){0, getb[mask] <= k, 0};
  if (!limit && !zero && dp[pos][mask].vis) return dp[pos][mask];
  int end = limit ? bit[pos] : 9;
  node ans = {1, 0, 0};
  for (int i = 0; i <= end; i++) {
    int nmask = mask | 1 << i;
    if (zero && !i) nmask = 0;
    node res = dfs(pos - 1, nmask, zero && !i, limit && i == end);
    inc(ans.cnt, res.cnt);
    inc(ans.sum, (1ll * i * powt[pos] % MOD * res.cnt % MOD + res.sum) % MOD);
  }
  if (!zero && !limit) dp[pos][mask] = ans;
  return ans;
}
int solve(long long x) {
  int cnt = 0;
  while (x) {
    bit[cnt++] = x % 10;
    x /= 10;
  }
  return dfs(cnt - 1, 0, true, true).sum;
}
int main() {
  for (int i = 0; i < (1 << 10); i++) {
    int t = i;
    while (t) {
      t &= t - 1;
      getb[i]++;
    }
  }
  powt[0] = 1;
  for (int i = 1; i < 25; i++) powt[i] = 10ll * powt[i - 1] % MOD;
  scanf("%I64d %I64d %d", &l, &r, &k);
  printf("%d\n", (solve(r) + MOD - solve(l - 1)) % MOD);
  return 0;
}
