#include <bits/stdc++.h>
using namespace std;
const int maxn = 999;
const int mod = 998244353;
long long dp1[maxn][2000];
long long cnt[maxn][2222];
long long val[maxn];
struct node {
  long long cnt, sum;
  node() {}
  node(long long c, long long s) {
    cnt = c;
    sum = s;
  }
};
long long f[maxn];
void init() {
  f[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    f[i] = (f[i - 1] * 10) % mod;
  }
}
long long kk;
node dfs(long long pos, long long sta, long long limit) {
  if (pos == -1) {
    long long cnt1 = 0;
    while (sta) {
      if (sta & 1) cnt1++;
      sta /= 2;
    }
    return node(cnt1 <= kk, 0);
  }
  if (cnt[pos][sta] != -1 && !limit) return node(cnt[pos][sta], dp1[pos][sta]);
  long long up = limit ? val[pos] : 9;
  long long nowcnt = 0;
  long long nowsum = 0;
  for (long long i = 0; i <= up; i++) {
    long long newsta = sta;
    if (i != 0 || newsta != 0) newsta |= (1 << i);
    node o = dfs(pos - 1, newsta, limit && i == up);
    nowcnt += o.cnt;
    nowsum += o.sum + ((f[pos] * i) % mod * o.cnt) % mod;
    nowsum %= mod;
    nowcnt %= mod;
  }
  if (!limit) {
    cnt[pos][sta] = nowcnt;
    dp1[pos][sta] = nowsum;
  }
  return node(nowcnt, nowsum);
}
void solve() {
  memset(cnt, -1, sizeof(cnt));
  long long l, r;
  scanf("%lld %lld %lld", &l, &r, &kk);
  l--;
  init();
  long long arr = -1;
  while (r) {
    val[++arr] = r % 10;
    r /= 10;
  }
  if (arr == -1) val[++arr] = 0;
  long long ans = 0;
  node aa = dfs(arr, 0, 1);
  ans += aa.sum;
  arr = -1;
  while (l) {
    val[++arr] = l % 10;
    l /= 10;
  }
  if (arr == -1) val[++arr] = 0;
  aa = dfs(arr, 0, 1);
  ans -= aa.sum;
  ans %= mod;
  ans += mod * 2;
  ans %= mod;
  printf("%lld\n", ans);
}
signed main() { solve(); }

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 999;
const int mod = 998244353;
long long dp1[maxn][2000];
long long cnt[maxn][2222];
long long val[maxn];
struct node {
  long long cnt, sum;
  node() {}
  node(long long c, long long s) {
    cnt = c;
    sum = s;
  }
};
long long f[maxn];
void init() {
  f[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    f[i] = (f[i - 1] * 10) % mod;
  }
}
long long kk;
node dfs(long long pos, long long sta, long long limit) {
  if (pos == -1) {
    long long cnt1 = 0;
    while (sta) {
      if (sta & 1) cnt1++;
      sta /= 2;
    }
    return node(cnt1 <= kk, 0);
  }
  if (cnt[pos][sta] != -1 && !limit) return node(cnt[pos][sta], dp1[pos][sta]);
  long long up = limit ? val[pos] : 9;
  long long nowcnt = 0;
  long long nowsum = 0;
  for (long long i = 0; i <= up; i++) {
    long long newsta = sta;
    if (i != 0 || newsta != 0) newsta |= (1 << i);
    node o = dfs(pos - 1, newsta, limit && i == up);
    nowcnt += o.cnt;
    nowsum += o.sum + ((f[pos] * i) % mod * o.cnt) % mod;
    nowsum %= mod;
    nowcnt %= mod;
  }
  if (!limit) {
    cnt[pos][sta] = nowcnt;
    dp1[pos][sta] = nowsum;
  }
  return node(nowcnt, nowsum);
}
void solve() {
  memset(cnt, -1, sizeof(cnt));
  long long l, r;
  scanf("%lld %lld %lld", &l, &r, &kk);
  l--;
  init();
  long long arr = -1;
  while (r) {
    val[++arr] = r % 10;
    r /= 10;
  }
  if (arr == -1) val[++arr] = 0;
  long long ans = 0;
  node aa = dfs(arr, 0, 1);
  ans += aa.sum;
  arr = -1;
  while (l) {
    val[++arr] = l % 10;
    l /= 10;
  }
  if (arr == -1) val[++arr] = 0;
  aa = dfs(arr, 0, 1);
  ans -= aa.sum;
  ans %= mod;
  ans += mod * 2;
  ans %= mod;
  printf("%lld\n", ans);
}
signed main() { solve(); }

```