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(chrono::steady_clock::now().time_since_epoch().count());
const int N = 20, K = 12;
const int M = 998244353;
int k;
int dp[N][(1 << K)];
int dpq[N][(1 << K)];
int pp(long long x) {
  vector<int> v;
  while (x) {
    v.push_back(x % 10);
    x /= 10;
  }
  reverse(v.begin(), v.end());
  memset(dp, 0, sizeof dp);
  memset(dpq, 0, sizeof dpq);
  for (int i = 0; i < v.size(); ++i) {
    for (int x = 0; x < (1 << 10); ++x) {
      int qx = 0;
      for (int j = 0; j < 10; ++j) {
        if ((x & (1 << j))) ++qx;
      }
      if (qx > k) continue;
      for (int j = 0; j < 10; ++j) {
        if (((x & (1 << j)) && qx == k) || qx < k) {
          dp[i + 1][(x | (1 << j))] =
              (dp[i + 1][(x | (1 << j))] +
               ((dp[i][x] * 10LL) % M + (dpq[i][x] * 1LL * j) % M) % M) %
              M;
          dpq[i + 1][(x | (1 << j))] =
              (dpq[i + 1][(x | (1 << j))] + dpq[i][x]) % M;
        }
      }
    }
    if (i == 0) {
      for (int j = 1; j < v[i]; ++j) {
        set<int> s;
        int x = 0;
        long long y = 0;
        for (int ii = 0; ii < i; ++ii) {
          s.insert(v[ii]);
          x |= (1 << v[ii]);
          y = (y * 10) + v[ii];
        }
        s.insert(j);
        x |= (1 << j);
        y = (y * 10) + j;
        if (s.size() <= k) {
          dp[i + 1][x] = (dp[i + 1][x] + (y % M)) % M;
          dpq[i + 1][x] = (dpq[i + 1][x] + 1) % M;
        }
      }
    } else {
      for (int j = 0; j < v[i]; ++j) {
        set<int> s;
        int x = 0;
        long long y = 0;
        for (int ii = 0; ii < i; ++ii) {
          s.insert(v[ii]);
          x |= (1 << v[ii]);
          y = (y * 10) + v[ii];
        }
        s.insert(j);
        x |= (1 << j);
        y = (y * 10) + j;
        if (s.size() <= k) {
          dp[i + 1][x] = (dp[i + 1][x] + (y % M)) % M;
          dpq[i + 1][x] = (dpq[i + 1][x] + 1) % M;
        }
      }
    }
  }
  int ans = 0;
  for (int x = 0; x < (1 << 10); ++x) ans = (ans + dp[v.size()][x]) % M;
  set<int> s;
  long long y = 0;
  for (int i = 0; i < v.size(); ++i) {
    s.insert(v[i]);
    y = (y * 10) + v[i];
  }
  if (s.size() <= k) ans = (ans + y) % M;
  return ans;
}
int p(long long x) {
  if (x == 0) return x;
  long long y = x;
  int q = 0;
  while (y) {
    ++q;
    y /= 10;
  }
  int ans = 0;
  y = 0;
  for (int i = 1; i < q; ++i) {
    y = (y * 10) + 9;
    ans = (ans + pp(y)) % M;
  }
  ans = (ans + pp(x)) % M;
  return ans;
}
int main() {
  long long l, r;
  scanf("%lld%lld%d", &l, &r, &k);
  printf("%d\n", (p(r) - p(l - 1) + M) % M);
  return 0;
}
