description: This version of the problem differs from the next one only in the constraint on n.
  
  Note that the memory limit in this problem is lower than in others.
  
  You have a vertical strip with n cells, numbered consecutively from 1 to n from top to bottom.
  
  You also have a token that is initially placed in cell n. You will move the token up until it arrives at cell 1.
  
  Let the token be in cell x > 1 at some moment. One shift of the token can have either of the following kinds: 
  
    * Subtraction: you choose an integer y between 1 and x-1, inclusive, and move the token from cell x to cell x - y. 
    * Floored division: you choose an integer z between 2 and x, inclusive, and move the token from cell x to cell ⌊ x/z ⌋ (x divided by z rounded down). 
  
  
  
  Find the number of ways to move the token from cell n to cell 1 using one or more shifts, and print it modulo m. Note that if there are several ways to move the token from one cell to another in one shift, all these ways are considered distinct (check example explanation for a better understanding).
  
  Input
  
  The only line contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 10^8 < m < 10^9; m is a prime number) — the length of the strip and the modulo.
  
  Output
  
  Print the number of ways to move the token from cell n to cell 1, modulo m.
  
  Examples
  
  Input
  
  
  3 998244353
  
  
  Output
  
  
  5
  
  
  Input
  
  
  5 998244353
  
  
  Output
  
  
  25
  
  
  Input
  
  
  42 998244353
  
  
  Output
  
  
  793019428
  
  Note
  
  In the first test, there are three ways to move the token from cell 3 to cell 1 in one shift: using subtraction of y = 2, or using division by z = 2 or z = 3.
  
  There are also two ways to move the token from cell 3 to cell 1 via cell 2: first subtract y = 1, and then either subtract y = 1 again or divide by z = 2.
  
  Therefore, there are five ways in total.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int maxn = 2e5 + 7;
  int n, m;
  long long f[maxn], s[maxn], d[maxn];
  signed main() {
    cin.sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> m;
    f[1] = 1;
    for (int i = 1; i <= n; ++i) {
      d[i] = (d[i] + d[i - 1]) % m;
      f[i] += (s[i - 1] + d[i]) % m;
      s[i] = (s[i - 1] + f[i]) % m;
      for (long long j = 2; j * i <= n; ++j) {
        d[i * j] = (d[i * j] + f[i]) % m;
        d[min(i * j + j, 1LL * n + 1)] =
            (d[min(i * j + j, 1LL * n + 1)] - f[i] + m) % m;
      }
    }
    cout << f[n] << '\n';
    return 0;
  }
  