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 node {
  long long cnt, sum;
  node(long long cnt = -1, long long sum = 0) : cnt(cnt), sum(sum) {}
} dp[20][2000];
int k;
long long cnt, sum;
int a[20];
long long ten[20];
int one(int x) {
  int cnt = 0;
  while (x) {
    cnt += x % 2;
    x >>= 1;
  }
  return cnt;
}
long long dfs(int pos, long long &cnt, int state, bool lead, bool limit) {
  if (pos == 0) {
    if (k < one(state)) return 0;
    cnt = 1;
    return 0;
  }
  if (!limit && !lead && dp[pos][state].cnt != -1) {
    cnt = dp[pos][state].cnt;
    return dp[pos][state].sum;
  }
  int up = limit ? a[pos] : 9;
  long long sum = 0;
  for (int i = 0; i <= up; i++) {
    long long tmp = 0;
    long long cntt = 0;
    if (lead && i == 0)
      tmp = dfs(pos - 1, cntt, state, lead && i == 0, limit && i == a[pos]);
    else
      tmp = dfs(pos - 1, cntt, state | (1 << i), lead && i == 0,
                limit && i == a[pos]);
    tmp %= mod;
    sum += (tmp + (1ll * i * ten[pos - 1] % mod * cntt % mod)) % mod;
    sum %= mod;
    cnt = (cnt + cntt) % mod;
  }
  if (!limit && !lead) dp[pos][state] = node(cnt, sum);
  return sum;
}
long long solve(long long x) {
  int pos = 0;
  while (x) {
    a[++pos] = x % 10;
    x /= 10;
  }
  long long t = 0;
  return dfs(pos, t, 0, true, true);
}
int main() {
  ten[0] = 1;
  for (int i = 1; i <= 19; i++) ten[i] = 10 * ten[i - 1] % mod;
  for (int i = 0; i <= 19; i++) {
    for (int j = 0; j <= 1024; j++) {
      dp[i][j].cnt = -1;
      dp[i][j].sum = 0;
    }
  }
  long long le, ri;
  scanf("%lld %lld %d", &le, &ri, &k);
  long long ans = solve(ri) - solve(le - 1);
  ans = (ans % mod + mod) % mod;
  printf("%lld\n", ans);
}
