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;
mt19937 rnd(51);
const long long MOD = 998244353;
const long long N = 30, MASK = (1 << 10) + 10;
long long k, n;
long long dp[N][2][MASK], ans[N][2][MASK];
string s;
vector<long long> num(18);
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    long long g = binpow(a, n / 2);
    return ((g * g) % MOD);
  } else {
    return ((binpow(a, n - 1) * a) % MOD);
  }
}
long long solve(long long i, long long f, long long msk, bool non_zero) {
  if (dp[i][f][msk] != -1) {
    return dp[i][f][msk];
  }
  dp[i][f][msk] = 0;
  if (!f) {
    for (long long j = 0; j < 10; j++) {
      if (msk & (1 << j)) {
        ans[i][0][msk] = (ans[i][0][msk] +
                          (solve(i + 1, 0, msk, 0) *
                           ((j * binpow(10, n - i - 1)) % MOD) % MOD) +
                          (solve(i + 1, 0, (msk ^ (1 << j)), 0) *
                           ((j * binpow(10, n - i - 1)) % MOD)) %
                              MOD +
                          ans[i + 1][0][msk] + ans[i + 1][0][msk ^ (1 << j)]) %
                         MOD;
        dp[i][0][msk] = (dp[i][0][msk] + solve(i + 1, 0, msk, 0) +
                         solve(i + 1, 0, (msk ^ (1 << j)), 0)) %
                        MOD;
      }
    }
  } else {
    for (long long j = 0; j < s[i] - '0'; j++) {
      if (non_zero && j == 0) continue;
      if (msk & (1 << j)) {
        ans[i][1][msk] = (ans[i][1][msk] +
                          (solve(i + 1, 0, msk, 0) *
                           ((j * binpow(10, n - i - 1)) % MOD) % MOD) +
                          (solve(i + 1, 0, (msk ^ (1 << j)), 0) *
                           ((j * binpow(10, n - i - 1)) % MOD)) %
                              MOD +
                          ans[i + 1][0][msk] + ans[i + 1][0][msk ^ (1 << j)]) %
                         MOD;
        dp[i][1][msk] = (dp[i][1][msk] + solve(i + 1, 0, msk, 0) +
                         solve(i + 1, 0, (msk ^ (1 << j)), 0)) %
                        MOD;
      }
    }
    long long j = s[i] - '0';
    if (msk & (1 << j)) {
      ans[i][1][msk] = (ans[i][1][msk] +
                        (solve(i + 1, 1, msk, 0) *
                         ((j * binpow(10, n - i - 1)) % MOD) % MOD) +
                        (solve(i + 1, 1, (msk ^ (1 << j)), 0) *
                         ((j * binpow(10, n - i - 1)) % MOD)) %
                            MOD +
                        ans[i + 1][1][msk] + ans[i + 1][1][msk ^ (1 << j)]) %
                       MOD;
      dp[i][1][msk] = (dp[i][1][msk] + solve(i + 1, 1, msk, 0) +
                       solve(i + 1, 1, (msk ^ (1 << j)), 0)) %
                      MOD;
    }
  }
  return dp[i][f][msk];
}
long long gett(long long v) {
  s = to_string(v);
  n = s.size();
  long long anss = 0;
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < 2; j++) {
      for (long long k = 0; k < MASK; k++) {
        dp[i][j][k] = -1;
        ans[i][j][k] = 0;
      }
    }
  }
  for (long long k = 0; k < MASK; k++) {
    dp[n][0][k] = dp[n][1][k] = 0;
  }
  dp[n][0][0] = dp[n][1][0] = 1;
  for (long long f = 0; f < MASK; f++) {
    solve(0, 1, f, 1);
  }
  for (long long f = 0; f < MASK; f++) {
    if (__builtin_popcountll(f) <= k) {
      anss = (anss + ans[0][1][f]) % MOD;
    }
  }
  return anss;
}
long long get(long long v) {
  if (v == 0) {
    return 0;
  }
  long long anss = 0;
  for (long long i = 1; i < to_string(v).size(); i++) {
    anss = (anss + gett(num[i - 1])) % MOD;
  }
  s = to_string(v);
  n = s.size();
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < 2; j++) {
      for (long long k = 0; k < MASK; k++) {
        dp[i][j][k] = -1;
        ans[i][j][k] = 0;
      }
    }
  }
  for (long long k = 0; k < MASK; k++) {
    dp[n][0][k] = dp[n][1][k] = 0;
  }
  dp[n][0][0] = dp[n][1][0] = 1;
  for (long long f = 0; f < MASK; f++) {
    solve(0, 1, f, 1);
  }
  for (long long f = 0; f < MASK; f++) {
    if (__builtin_popcountll(f) <= k) {
      anss = (anss + ans[0][1][f]) % MOD;
    }
  }
  return anss;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  num[0] = 9;
  for (long long i = 1; i < 17; i++) {
    num[i] = num[i - 1] * 10 + 9;
  }
  long long l, r;
  cin >> l >> r >> k;
  cout << ((get(r) - get(l - 1) + MOD) % MOD) << endl;
  return 0;
}
