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 INF = 0x3f3f3f3;
const long long mo = 998244353;
long long add(long long a, long long b) { return (a + b) % mo; }
long long mul(long long a, long long b) { return (1ll * a * b) % mo; }
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long l, r, k, mi[20];
pair<long long, long long> dp[20][1 << 10];
long long num[20];
pair<long long, long long> cnt(long long tot, long long is, long long S) {
  if (!tot) return make_pair(1, 0);
  if (!is && ~dp[tot][S].first) return dp[tot][S];
  pair<long long, long long> ret = make_pair(0, 0);
  long long i;
  for (long long i = is ? num[tot] : 9; ~i; i--) {
    long long nxt = (S | i) ? (S | 1 << i) : 0;
    if (__builtin_popcount(nxt) > k) continue;
    pair<long long, long long> ntx = cnt(tot - 1, is && i == num[tot], nxt);
    ret.first = (ret.first + ntx.first) % mo;
    ret.second = (ret.second +
                  (ntx.second + 1ll * ntx.first * mi[tot - 1] % mo * i) % mo) %
                 mo;
  }
  if (!is) dp[tot][S] = ret;
  return ret;
}
long long solve(long long cnter) {
  if (cnter < 1) return 0ll;
  long long tot = 0;
  while (cnter) {
    num[++tot] = cnter % 10;
    cnter /= 10;
  }
  return cnt(tot, 1, 0).second;
}
signed main() {
  __builtin_memset(dp, -1, sizeof(dp));
  long long i;
  for (i = 0; i <= 19; i += 1) mi[i] = !i ? 1 : mi[i - 1] * 10 % mo;
  l = read(), r = read(), k = read();
  printf("%lld", ((solve(r) - solve(l - 1)) % mo + mo) % mo);
  return 0;
}
