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;
void read(long long &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  x *= f;
  return;
}
long long fs(long long x) { return x < 0 ? -x : x; }
void print(long long x) {
  if (x < 10) {
    putchar(x + '0');
    return;
  }
  print(x / 10);
  putchar(x % 10 + '0');
  return;
}
void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = fs(x);
  }
  print(x);
  return;
}
long long mint(long long x, long long y) { return x < y ? x : y; }
long long maxt(long long x, long long y) { return x > y ? x : y; }
long long l, r, k, pow_[10 << 1], f[10 << 1][1 << 10][2][2],
    g[10 << 1][1 << 10][2][2], a[10 << 1], bit[1 << 10];
void dfs(long long pos, long long s, bool qd, bool sx) {
  if (pos == 0) {
    f[pos][s][qd][sx] = 0;
    g[pos][s][qd][sx] = bit[s] <= k;
    return;
  }
  if (~g[pos][s][qd][sx]) return;
  long long up = sx ? a[pos] : 9;
  f[pos][s][qd][sx] = g[pos][s][qd][sx] = 0;
  for (int i = 0; i <= up; i++) {
    long long ns = (qd && (!i)) ? s : (s | (1 << i));
    dfs(pos - 1, ns, qd && (!i), sx && i == up);
    g[pos][s][qd][sx] =
        (g[pos][s][qd][sx] + g[pos - 1][ns][qd && (!i)][sx && i == up]) %
        998244353;
    f[pos][s][qd][sx] =
        (f[pos][s][qd][sx] + (f[pos - 1][ns][qd && (!i)][sx && i == up] +
                              (g[pos - 1][ns][qd && (!i)][sx && i == up] *
                               (pow_[pos - 1] * i) % 998244353) %
                                  998244353) %
                                 998244353) %
        998244353;
  }
  return;
}
long long solve(long long x) {
  memset(f, -1, sizeof(f));
  memset(g, -1, sizeof(g));
  memset(a, 0, sizeof(a));
  long long len = 0;
  while (x) {
    a[++len] = x % 10;
    x /= 10;
  }
  dfs(len, 0, 1, 1);
  return f[len][0][1][1];
}
int main() {
  pow_[0] = 1, bit[0] = 0;
  for (int i = 1; i < 1 << 10; i++)
    if (i & 1)
      bit[i] = bit[i >> 1] + 1;
    else
      bit[i] = bit[i >> 1];
  for (int i = 1; i <= 18; i++)
    pow_[i] =
        ((pow_[i - 1] << 3) % 998244353 + (pow_[i - 1] << 1) % 998244353) %
        998244353;
  read(l), read(r), read(k);
  printf("%lld", (solve(r) - solve(l - 1) + 998244353) % 998244353);
}
