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 dv = 998244353;
long long d, a[18], dp[2][2][2][2][1ll << 10];
long long val(long long x) {
  long long i, j, ii, jj, mk, *tg, *tg2, ml = 1e18, z = 0;
  for (i = 17; i + 1; i--) {
    a[i] = x % 10;
    x /= 10;
  }
  dp[0][0][0][0][0] = 1;
  for (i = 0; i < 18; i++) {
    ml /= 10;
    for (ii = 0; ii < 2; ii++) {
      for (jj = 0; jj < 2; jj++) {
        for (mk = 0; mk < 1ll << 10; mk++) {
          for (j = 0; j <= (!jj ? a[i] : 9); j++) {
            tg = &dp[0][1][ii || j][jj || j < a[i]][mk | (j || ii) << j];
            tg2 = &dp[1][1][ii || j][jj || j < a[i]][mk | (j || ii) << j];
            *tg = (*tg + dp[0][0][ii][jj][mk]) % dv;
            *tg2 = (*tg2 + dp[1][0][ii][jj][mk] +
                    j * ml % dv * dp[0][0][ii][jj][mk]) %
                   dv;
          }
        }
      }
    }
    for (ii = 0; ii < 2; ii++) {
      for (jj = 0; jj < 2; jj++) {
        for (mk = 0; mk < 1ll << 10; mk++) {
          dp[0][0][ii][jj][mk] = dp[0][1][ii][jj][mk];
          dp[1][0][ii][jj][mk] = dp[1][1][ii][jj][mk];
          dp[0][1][ii][jj][mk] = 0;
          dp[1][1][ii][jj][mk] = 0;
        }
      }
    }
  }
  for (ii = 0; ii < 2; ii++) {
    for (jj = 0; jj < 2; jj++) {
      for (mk = 0; mk < 1ll << 10; mk++) {
        if (__builtin_popcountll(mk) <= d) {
          z = (z + dp[1][0][ii][jj][mk]) % dv;
          dp[0][0][ii][jj][mk] = 0;
          dp[1][0][ii][jj][mk] = 0;
        }
      }
    }
  }
  return z;
}
int main() {
  long long k, l;
  scanf("%lld%lld%lld", &k, &l, &d);
  printf("%lld\n", (val(l) + dv - val(k - 1)) % dv);
}
