Problem: Let's define the function f of multiset a as the multiset of number of occurences of every number, that is present in a.

E.g., f(\{5, 5, 1, 2, 5, 2, 3, 3, 9, 5\}) = \{1, 1, 2, 2, 4\}.

Let's define f^k(a), as applying f to array a k times: f^k(a) = f(f^{k-1}(a)), f^0(a) = a. 

E.g., f^2(\{5, 5, 1, 2, 5, 2, 3, 3, 9, 5\}) = \{1, 2, 2\}.

You are given integers n, k and you are asked how many different values the function f^k(a) can have, where a is arbitrary non-empty array with numbers of size no more than n. Print the answer modulo 998 244 353.

Input

The first and only line of input consists of two integers n, k (1 ≤ n, k ≤ 2020).

Output

Print one number — the number of different values of function f^k(a) on all possible non-empty arrays with no more than n elements modulo 998 244 353.

Examples

Input


3 1


Output


6


Input


5 6


Output


1


Input


10 1


Output


138


Input


10 2


Output


33

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int cys = 998244353;
int n, k;
long long mod(long long x) { return x >= cys ? x - cys : x; }
namespace task1 {
long long d[2025][2025];
void work() {
  d[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      d[i][j] = d[i - 1][j];
      if (j >= i) d[i][j] = mod(d[i][j] + d[i][j - i]);
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = mod(ans + d[n][i]);
  printf("%lld\n", ans);
}
}  // namespace task1
namespace task2 {
long long d[2025][2025], f[2025][2025];
void work() {
  d[0][0] = 1;
  for (int i = n; i >= 1; i--)
    for (int j = 0; j <= n / i; j++)
      for (int k = 0; k <= n; k++)
        if (j && k >= i * j) d[j][k] = mod(d[j][k] + d[j - 1][k - i * j]);
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= n; j++) ans = mod(ans + d[i][j]);
  printf("%lld\n", ans);
}
}  // namespace task2
namespace taskg3 {
vector<int> a;
map<vector<int>, bool> bk;
int ans;
bool check(vector<int> x) {
  vector<int> tmp(0);
  for (int i = 0; i < x.size(); i++)
    for (int j = 0; j < x[i]; j++) tmp.push_back(i + 1);
  int sum = 0;
  for (int i = 0; i < tmp.size(); i++) sum += tmp[i] * (tmp.size() - i);
  return sum <= n;
}
vector<int> getans(vector<int> x) {
  sort(x.begin(), x.end());
  int lst = 0;
  vector<int> ret(0);
  for (int i = 1; i < x.size(); i++)
    if (x[i] != x[i - 1]) ret.push_back(i - lst), lst = i;
  ret.push_back(x.size() - lst);
  return ret;
}
bool dfs(int now, int sum) {
  if (now >= 1) {
    if (!check(a)) return 0;
    vector<int> tmp = a;
    for (int i = 1; i <= k - 3; i++) tmp = getans(tmp);
    sort(tmp.begin(), tmp.end());
    if (!bk.count(tmp)) bk[tmp] = 1, ans++;
  }
  int g = now ? a[now - 1] : 64;
  for (int i = 1; i <= g && i <= 64 - sum; i++) {
    a.push_back(i);
    if (!dfs(now + 1, sum + i)) {
      a.pop_back();
      break;
    }
    a.pop_back();
  }
  return 1;
}
void work() {
  ans = 0;
  dfs(0, 0);
  printf("%d\n", ans);
}
}  // namespace taskg3
int main() {
  n = readint();
  k = readint();
  if (k >= 20)
    printf("1\n");
  else if (k == 1)
    task1::work();
  else if (k == 2)
    task2::work();
  else
    taskg3::work();
  return 0;
}