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 int maxn = 1e6 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const int mod = 998244353;
long long l, r, k;
long long dp[25][1058][3], dp1[25][1058][3], pows[maxn], digit[maxn];
bool valid(long long x) {
  long long cnt = 0;
  while (x) cnt += x % 2, x /= 2;
  if (cnt <= k)
    return 1;
  else
    return 0;
}
long long cnt(long long x) {
  memset(dp, 0, sizeof(dp));
  memset(dp1, 0, sizeof(dp1));
  long long cnt = 0;
  if (x == 0) return 0;
  while (x != 0) {
    digit[cnt++] = x % 10;
    x /= 10;
  }
  reverse(digit, digit + cnt);
  for (long long i = 1; i <= 9; i++) {
    if (i < digit[0])
      dp[0][1 << i][0] = (pows[cnt - 1] * i) % mod, dp1[0][1 << i][0] = 1;
    else if (i == digit[0])
      dp[0][1 << i][1] = (pows[cnt - 1] * i) % mod, dp1[0][1 << i][1] = 1;
  }
  for (long long i = 0; i < cnt; i++) dp1[i][0][0] = 1;
  for (long long i = 0; i < cnt - 1; i++) {
    for (long long j = 0; j < (1 << 10); j++) {
      if (j == 1) continue;
      for (long long k = 0; k < 10; k++) {
        if (valid(j | (1 << k)) && (j | (1 << k)) != 1) {
          dp[i + 1][(j | (1 << k))][0] =
              (dp[i + 1][(j | (1 << k))][0] + dp[i][j][0]) % mod;
          dp1[i + 1][(j | (1 << k))][0] =
              (dp1[i + 1][(j | (1 << k))][0] + dp1[i][j][0]) % mod;
          dp[i + 1][(j | (1 << k))][0] =
              (dp[i + 1][(j | (1 << k))][0] +
               pows[cnt - i - 2] * k % mod * dp1[i][j][0]) %
              mod;
          if (k == digit[i + 1]) {
            dp[i + 1][(j | (1 << k))][1] =
                (dp[i + 1][(j | (1 << k))][1] + dp[i][j][1]) % mod;
            dp1[i + 1][(j | (1 << k))][1] =
                (dp1[i + 1][(j | (1 << k))][1] + dp1[i][j][1]) % mod;
            dp[i + 1][(j | (1 << k))][1] =
                (dp[i + 1][(j | (1 << k))][1] +
                 pows[cnt - i - 2] * k % mod * dp1[i][j][1]) %
                mod;
          } else if (k < digit[i + 1]) {
            dp[i + 1][(j | (1 << k))][0] =
                (dp[i + 1][(j | (1 << k))][0] + dp[i][j][1]) % mod;
            dp1[i + 1][(j | (1 << k))][0] =
                (dp1[i + 1][(j | (1 << k))][0] + dp1[i][j][1]) % mod;
            dp[i + 1][(j | (1 << k))][0] =
                (dp[i + 1][(j | (1 << k))][0] +
                 pows[cnt - i - 2] * k % mod * dp1[i][j][1]) %
                mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (long long j = 2; j < (1 << 10); j++) {
    ans = (ans + dp[cnt - 1][j][0]) % mod;
    ans = (ans + dp[cnt - 1][j][1]) % mod;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pows[0] = 1;
  for (long long i = 1; i <= 20; i++) {
    pows[i] = (pows[i - 1] * 10) % mod;
  }
  cin >> l >> r >> k;
  cout << (cnt(r) + mod - cnt(l - 1)) % mod << endl;
}
