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>
using namespace std;
const long long mod = 998244353;
struct Res {
  long long sum, cnt;
} dp[21][2000];
Res tp;
long long ten[101];
int a[21];
int K;
Res dfs(int pos, int sta, bool limit, bool lead) {
  if (__builtin_popcount(sta) > K) {
    tp.sum = tp.cnt = 0;
    return tp;
  }
  if (pos == -1) {
    tp.sum = 0;
    tp.cnt = 1;
    return tp;
  }
  if (!limit && !lead && dp[pos][sta].cnt != -1) return dp[pos][sta];
  int up = limit ? a[pos] : 9;
  long long ans = 0, cnt = 0;
  for (int i = 0; i <= up; i++) {
    if (sta & (1 << i)) {
      tp = dfs(pos - 1, sta, limit && i == up, lead && i == 0);
      cnt += tp.cnt;
      ans = (ans + tp.sum + tp.cnt % mod * i % mod * ten[pos] % mod) % mod;
    } else {
      if (lead && i == 0) {
        tp = dfs(pos - 1, sta, limit && i == up, 1);
        cnt += tp.cnt;
        ans = (ans + tp.sum + tp.cnt % mod * i % mod * ten[pos] % mod) % mod;
      } else {
        tp = dfs(pos - 1, sta | (1 << i), limit && i == up, 0);
        cnt += tp.cnt;
        ans = (ans + tp.sum + tp.cnt % mod * i % mod * ten[pos] % mod) % mod;
      }
    }
  }
  tp.sum = ans;
  tp.cnt = cnt;
  if (!limit && !lead) dp[pos][sta] = tp;
  return tp;
}
long long solve(long long x) {
  int pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, 1, 1).sum;
}
int main() {
  ten[0] = 1;
  for (int i = 1; i <= 100; i++) ten[i] = ten[i - 1] * 10 % mod;
  long long x, y;
  while (~scanf("%lld%lld%d", &x, &y, &K)) {
    memset(dp, -1, sizeof(dp));
    printf("%lld\n", (solve(y) - solve(x - 1) + mod) % mod);
  }
  return 0;
}
