Given is an integer N. How many permutations (P_0,P_1,\cdots,P_{2N-1}) of (0,1,\cdots,2N-1) satisfy the following condition?

* For each i (0 \leq i \leq 2N-1), N^2 \leq i^2+P_i^2 \leq (2N)^2 holds.



Since the number can be enormous, compute it modulo M.

Constraints

* 1 \leq N \leq 250
* 2 \leq M \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M


Output

Print the number of permutations that satisfy the condition, modulo M.

Examples

Input

2 998244353


Output

4


Input

10 998244353


Output

53999264


Input

200 998244353


Output

112633322
/*
Author: QAQAutomaton
Lang: C++
Code: F.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef complex<double> cp;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T,class T2>int chkmin(T &a,T2 b){return a>b?a=b,1:0;}
template<class T,class T2>int chkmax(T &a,T2 b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T,class T2>T mmin(T a,T2 b){return a<b?a:b;}
template<class T,class T2>T mmax(T a,T2 b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		memset(&inf,0x3f,sizeof(inf));
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;

int n,p;
int xl[505],xr[505];
int yl[505],yr[505];
int f[505][505];
int las[505];
signed main(){
#ifdef QAQAutoMaton 
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	read(n,p);
	for(int i=0;i<n;++i){
		int l=0,r=n+n-1;
		while(l*l+i*i<n*n)++l;
		while(r*r+i*i>4*n*n)--r;
		xl[i+1]=l;xr[i+1]=r;
	}
	for(int i=n;i<n+n;++i){
		int l=0,r=n+n-1;
		while(l*l+i*i<n*n)++l;
		while(r*r+i*i>4*n*n)--r;
		yl[i-n+1]=l;yr[i-n+1]=r;
	}
	for(int i=1;i<=n;++i)++xr[i],++yr[i];
	f[0][0]=1;
	reverse(xl+1,xl+n+1);
	reverse(yl+1,yl+n+1);
	reverse(xr+1,xr+n+1);
	reverse(yr+1,yr+n+1);
	int ans=0;
	for(int i=1;i<=n;++i){
		las[i]=las[i-1];
		while(yr[las[i]+1]<=xl[i])++las[i];
	}
	las[n+1]=n;
	for(int k=0;k<=n;++k){
		f[0][0]=1;	
		for(int k=las[0]+1;k<=las[1];++k){
			f[0][0]=f[0][0]*(yr[k]-k+1)%p;
		}
		for(int i=1;i<=n;++i){
			for(int j=0;j<=i;++j){
				f[i][j]=(f[i-1][j]*(xr[i]-i+1-n+j-k)+(j?f[i-1][j-1]*(xl[i]-(j-1+las[i])):0))%p;
			}
			for(int j=0;j<=i;++j)
				for(int k=las[i]+1;k<=las[i+1];++k){
					f[i][j]=f[i][j]*(yr[k]-j-k+1)%p;
				}
		}
		(ans+=(k&1?p-f[n][k]:f[n][k]))>=p?ans-=p:0;
	}
	write(ans,'\n');
//	for(int i=1;i<=n;++i)
	return 0;
}

