Problem:
Score : 1200 points

Problem Statement

N problems have been chosen by the judges, now it's time to assign scores to them!

Problem i must get an integer score A_i between 1 and N, inclusive.
The problems have already been sorted by difficulty: A_1 \le A_2 \le \ldots \le A_N must hold.
Different problems can have the same score, though.

Being an ICPC fan, you want contestants who solve more problems to be ranked higher.
That's why, for any k (1 \le k \le N-1), you want the sum of scores of any k problems to be strictly less than the sum of scores of any k+1 problems.

How many ways to assign scores do you have? Find this number modulo the given prime M.

Constraints

2 \leq N \leq 5000

9 \times 10^8 < M < 10^9

M is a prime.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of ways to assign scores to the problems, modulo M.

Sample Input 1

2 998244353

Sample Output 1

3

The possible assignments are (1, 1), (1, 2), (2, 2).

Sample Input 2

3 998244353

Sample Output 2

7

The possible assignments are (1, 1, 1), (1, 2, 2), (1, 3, 3), (2, 2, 2), (2, 2, 3), (2, 3, 3), (3, 3, 3).

Sample Input 3

6 966666661

Sample Output 3

66

Sample Input 4

96 925309799

Sample Output 4

83779

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

const int maxn = 5010;

int n;
// dp[lst]
long long dp[maxn];
long long m;
long long arr[maxn];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    dp[0] = 1;
    for(int i = 1, j = n + 2; i <= j; i++, j--) {
        arr[i] = arr[j] = i - 1;
    }
    for(int i = 2; i <= n; i++) {
        for(int j = arr[i]; j <= n; j++) {
            dp[j] += dp[j - arr[i]];
            dp[j] %= m;
        }
    }
    long long ret = 0;
    for(int i = 0; i <= n; i++) {
        ret += i * dp[n - i];
        ret %= m;
    }
    cout << ret << endl;
}