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;
long long dp[20][1 << 10][2][2];
long long fp[20][1 << 10][2][2];
const long long MOD = 998244353;
void add(long long& x, const long long& y) { x = (x + y) % MOD; }
long long calc(long long x, int k) {
  memset(dp, 0, sizeof dp);
  memset(fp, 0, sizeof fp);
  for (int i = 0; i < 10; ++i) {
    dp[0][1 << i][0][i == 0] = 1;
    fp[0][1 << i][0][i == 0] = i;
    if (i <= x % 10) {
      dp[0][1 << i][1][i == 0] = 1;
      fp[0][1 << i][1][i == 0] = i;
    }
  }
  long long y = x / 10;
  long long tpow = 10;
  for (int i = 0; i < 18 && y; ++i, y /= 10, tpow = tpow * 10 % MOD) {
    for (int j = 0; j < (1 << 10); ++j) {
      for (int q = 0; q < 2; ++q) {
        for (int k = 0; k < 10; ++k) {
          int z = j | (1 << k);
          add(dp[i + 1][z][0][k == 0], dp[i][j][0][q]);
          add(fp[i + 1][z][0][k == 0],
              fp[i][j][0][q] + dp[i][j][0][q] * k * tpow);
          if (k < y % 10) {
            add(dp[i + 1][z][1][k == 0], dp[i][j][0][q]);
            add(fp[i + 1][z][1][k == 0],
                fp[i][j][0][q] + dp[i][j][0][q] * k * tpow);
          } else if (k == y % 10) {
            add(dp[i + 1][z][1][k == 0], dp[i][j][1][q]);
            add(fp[i + 1][z][1][k == 0],
                fp[i][j][1][q] + dp[i][j][1][q] * k * tpow);
          }
        }
      }
    }
  }
  y = x;
  long long ans = 0;
  for (int i = 0; y; ++i, y /= 10) {
    for (int j = 0; j < (1 << 10); ++j) {
      if (__builtin_popcount(j) > k) continue;
      if (y < 10) {
        add(ans, fp[i][j][1][0]);
      } else {
        add(ans, fp[i][j][0][0]);
      }
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long l, r, k;
  cin >> l >> r >> k;
  long long x = (calc(r, k) - calc(l - 1, k) + MOD) % MOD;
  cout << x << "\n";
  return 0;
}
