Problem: This is the hard version of the problem. The only difference between the easy version and the hard version is the constraints on n. You can only make hacks if both versions are solved.

A permutation of 1, 2, …, n is a sequence of n integers, where each integer from 1 to n appears exactly once. For example, [2,3,1,4] is a permutation of 1, 2, 3, 4, but [1,4,2,2] isn't because 2 appears twice in it.

Recall that the number of inversions in a permutation a_1, a_2, …, a_n is the number of pairs of indices (i, j) such that i < j and a_i > a_j.

Let p and q be two permutations of 1, 2, …, n. Find the number of permutation pairs (p,q) that satisfy the following conditions:

  * p is lexicographically smaller than q. 
  * the number of inversions in p is greater than the number of inversions in q. 



Print the number of such pairs modulo mod. Note that mod may not be a prime.

Input

The only line contains two integers n and mod (1≤ n≤ 500, 1≤ mod≤ 10^9).

Output

Print one integer, which is the answer modulo mod.

Example

Input


4 403458273


Output


17

Note

The following are all valid pairs (p,q) when n=4.

  * p=[1,3,4,2], q=[2,1,3,4], 
  * p=[1,4,2,3], q=[2,1,3,4], 
  * p=[1,4,3,2], q=[2,1,3,4], 
  * p=[1,4,3,2], q=[2,1,4,3], 
  * p=[1,4,3,2], q=[2,3,1,4], 
  * p=[1,4,3,2], q=[3,1,2,4], 
  * p=[2,3,4,1], q=[3,1,2,4], 
  * p=[2,4,1,3], q=[3,1,2,4], 
  * p=[2,4,3,1], q=[3,1,2,4], 
  * p=[2,4,3,1], q=[3,1,4,2], 
  * p=[2,4,3,1], q=[3,2,1,4], 
  * p=[2,4,3,1], q=[4,1,2,3], 
  * p=[3,2,4,1], q=[4,1,2,3], 
  * p=[3,4,1,2], q=[4,1,2,3], 
  * p=[3,4,2,1], q=[4,1,2,3], 
  * p=[3,4,2,1], q=[4,1,3,2], 
  * p=[3,4,2,1], q=[4,2,1,3]. 

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#include <unordered_set>
#include <unordered_map>
#define rep(x, s, t) for(llint x = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(llint x = (s); (x) >= (t); (x)--)
#define reps(x, s) for(llint x = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define outf(x) cout << fixed << setprecision(16) << (x) << endl
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

ll mod = 1;
//const ll mod = 1000000007;
//const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

vector<mint> fact, fact_inv;
void make_fact(int n){
	fact.resize(n+1), fact_inv.resize(n+1);
	fact[0] = mint(1); rep(i, 1, n) fact[i] = fact[i-1] * mint(i);
	fact_inv[n] = fact[n].inverse(); per(i, n-1, 0) fact_inv[i] = fact_inv[i+1] * mint(i+1);
}
mint comb(ll n, ll k){ if(n < 0 || k < 0 || n < k) return mint(0); return fact[n] * fact_inv[k] * fact_inv[n-k];}
mint perm(ll n, ll k){ return comb(n, k) * fact[k]; }

vector<int> prime;
void make_prime(int n){
	prime.resize(n+1);
	rep(i, 2, n){
		if(prime[i]) continue;
		for(int j = i; j <= n; j+=i) prime[j] = i;
	}
}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "YES" << endl; }
void no(){ cout << "NO" << endl; }
ll sgn(ll x){ if(x > 0) return 1; if(x < 0) return -1; return 0;}
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll lcm(ll a, ll b){return a/gcd(a, b)*b;}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}
string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;}
ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;}
template<typename T>
void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}

template<class S, class T> pair<S, T>& operator+=(pair<S,T> &s, const pair<S,T> &t){
	s.first += t.first, s.second += t.second;
	return s;
}
template<class S, class T> pair<S, T>& operator-=(pair<S,T> &s, const pair<S,T> &t){
	s.first -= t.first, s.second -= t.second;
	return s;
}
template<class S, class T> pair<S, T> operator+(const pair<S,T> &s, const pair<S,T> &t){
	return pair<S,T>(s.first+t.first, s.second+t.second);
}
template<class S, class T> pair<S, T> operator-(const pair<S,T> &s, const pair<S,T> &t){
	return pair<S,T>(s.first-t.first, s.second-t.second);
}
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++; if(itr != map_var.end()) os << ","; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, multiset<T>& set_var) {
	for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t){for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}cout << endl;}
void dump_func(){cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}


ll n, M;
mint dp[350005], ndp[350005], dpsum[350005];
ll zero = 175002;

mint calc(ll n)
{
	mint ret = mint(0);
	rep(i, 0, n-1) rep(j, 0, n-1){
		if(i >= j) continue;
		//outl(i, j, dpsum[M+zero] - dpsum[j-i+zero]);
		ret += dpsum[M+zero] - dpsum[j-i+zero];
	}
	//outa(dp, -M+zero, M+zero);
	
	rep(j, -M, M){
		ll l = max(j-(n-1), -M), r = j;
		ndp[j+zero] = dpsum[r+zero] - dpsum[l-1+zero];
	}
	rep(j, -M, M) dp[j+zero] = ndp[j+zero], dpsum[j+zero] = dpsum[j-1+zero] + dp[j+zero];
	
	rep(j, -M, M){
		ll l = j, r = min(j+(n-1), M);
		ndp[j+zero] = dpsum[r+zero] - dpsum[l-1+zero];
	}
	rep(j, -M, M) dp[j+zero] = ndp[j+zero], dpsum[j+zero] = dpsum[j-1+zero] + dp[j+zero];
	
	//outl(n, ret);
	return ret;
}

mint perm2(ll n, ll k)
{
	mint ret = mint(1);
	rep(i, 1, k) ret *= mint(n+1-i);
	return ret;
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> mod;
	if(mod == 1 || n == 1){
		outl(0);
		return 0;
	}
	M = n*(n+1)/2;
	//outl(M);
	
	//make_fact(1005);
	
	dp[zero] = mint(1);
	rep(j, -M, M) dpsum[j+zero] = dpsum[j-1+zero] + dp[j+zero];
	
	mint ans = mint(0);
	rep(i, 1, n) ans += calc(i) * perm2(n, n-i);
	outl(ans);
	
	return 0;
}