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:xht37
#include <bits/stdc++.h>

#define ui unsigned int
#define ll long long
#define ul unsigned ll
#define ld long double

#define pi pair <int, int>
#define fi first
#define se second
#define mp make_pair

#define ls (p << 1)
#define rs (ls | 1)
#define md ((t[p].l + t[p].r) >> 1)

#define vi vector <int>
#define pb push_back
#define pq priority_queue

#define dbg(x) cerr << #x" = " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)

#define fl(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)

using namespace std;

namespace io {
	const int SI = 1 << 21 | 1;
	char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
	int f, t;
	#define gc() (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin), IS == IT ? EOF : *IS++) : *IS++)
	inline void flush() {
		fwrite(OB, 1, OS - OB, stdout), OS = OB;
	}
	inline void pc(char x) {
		*OS++ = x;
		if (OS == OT) flush();
	}

	template <class I>
	inline void rd(I &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15), c = gc());
		x *= f;
	}
	template <class I>
	inline void rd(I &x, I &y) {
		rd(x), rd(y);
	}
	template <class I>
	inline void rd(I &x, I &y, I &z) {
		rd(x), rd(y), rd(z);
	}
	template <class I>
	inline void rda(I *a, int n) {
		for (int i = 1; i <= n; i++) rd(a[i]);
	}
	inline void rdc(char &c) {
		for (c = gc(); c < 33 || c > 126; c = gc());
	}
	inline void rds(char *s, int &n) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (n = 0; c >= 33 && c <= 126; s[++n] = c, c = gc());
		s[n+1] = '\0';
	}
	inline void rds(string &s) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (s.clear(); c >= 33 && c <= 126; s.pb(c), c = gc());
	}

	template <class I>
	inline void print(I x, char k = '\n') {
		if (!x) pc('0');
		if (x < 0) pc('-'), x = -x;
		while (x) ch[++t] = x % 10 + '0', x /= 10;
		while (t) pc(ch[t--]);
		pc(k);
	}
	template <class I>
	inline void print(I x, I y) {
		print(x, ' '), print(y);
	}
	template <class I>
	inline void print(I x, I y, I z) {
		print(x, ' '), print(y, ' '), print(z);
	}
	template <class I>
	inline void printa(I *a, int n) {
		for (int i = 1; i <= n; i++) print(a[i], " \n"[i==n]);
	}
	inline void printc(char c) {
		pc(c);
	}
	inline void prints(char *s, int n) {
		for (int i = 1; i <= n; i++) pc(s[i]);
		pc('\n');
	}
	inline void prints(string s) {
		int n = s.length();
		while (t < n) pc(s[t++]);
		pc('\n'), t = 0;
	}
	struct Flush {
		~Flush() {
			flush();
		}
	} flusher;
}
using io::rd;
using io::rda;
using io::rdc;
using io::rds;
using io::print;
using io::printa;
using io::printc;
using io::prints;

int P; 

struct modint {
    int x;
    inline modint(int x = 0) : x(x) {}
    inline modint &operator = (int o) { return x = o, *this; }
    inline modint &operator += (modint o) { return (x += o.x) >= P && (x -= P), *this; }
    inline modint &operator -= (modint o) { return (x -= o.x) < 0 && (x += P), *this; }
    inline modint &operator *= (modint o) { return x = 1ll * x * o.x % P, *this; }
    template <class I>
    inline modint &operator ^= (I b) {
        modint a = *this, c;
        if (!~b) b = P - 2;
        c.x = 1 % P;
        while (b) {
            if (b & 1) c *= a;
            a *= a, b >>= 1;
        }
        return x = c.x, *this;
    }
    inline modint &operator /= (modint o) { return *this *= o ^ -1; }
    inline modint &operator += (int o) { return (x += o) >= P && (x -= P), *this; }
    inline modint &operator -= (int o) { return (x -= o) < 0 && (x += P), *this; }
    inline modint &operator *= (int o) { return x = 1ll * x * o % P, *this; }
    inline modint &operator /= (int o) { return *this *= (modint)o ^ -1; }
    template <class I>
    inline friend modint operator + (modint a, I b) { return a += b; }
    template <class I>
    inline friend modint operator - (modint a, I b) { return a -= b; }
    template <class I>
    inline friend modint operator * (modint a, I b) { return a *= b; }
    template <class I>
    inline friend modint operator ^ (modint a, I b) { return a ^= b; }
    template <class I>
    inline friend modint operator / (modint a, I b) { return a /= b; }
    inline friend bool operator == (modint a, int b) { return a.x == b; }
    inline friend bool operator != (modint a, int b) { return a.x != b; }
    inline friend bool operator < (modint a, int b) { return a.x < b; }
    inline friend bool operator <= (modint a, int b) { return a.x <= b; }
    inline friend bool operator > (modint a, int b) { return a.x > b; }
    inline friend bool operator >= (modint a, int b) { return a.x >= b; }
    inline friend bool operator == (modint a, modint b) { return a.x == b.x; }
    inline friend bool operator != (modint a, modint b) { return a.x != b.x; }
    inline friend bool operator < (modint a, modint b) { return a.x < b.x; }
    inline friend bool operator <= (modint a, modint b) { return a.x <= b.x; }
    inline friend bool operator > (modint a, modint b) { return a.x > b.x; }
    inline friend bool operator >= (modint a, modint b) { return a.x >= b.x; }
    inline bool operator ! () { return !x; }
    inline modint operator - () { return x ? P - x : 0; }
};
inline void rd(modint &x) { rd(x.x); }
inline void print(modint x, char k = '\n') { print(x.x, k); }

//const int NP = 1e6 + 7;
//modint p[NP], v[NP], vp[NP];
//inline void init(int n) {
//  p[0] = v[0] = 1;
//  for (int i = 1; i <= n; i++) p[i] = p[i-1] * i;
//  vp[n] = 1 / p[n];
//  for (int i = n; i; i--) v[i] = vp[i] * p[i-1], vp[i-1] = vp[i] * i;
//}
//inline modint binom(int n, int m) {
//  return (m < 0 || n < m) ? 0 : p[n] * vp[m] * vp[n-m];
//}

const int N = 507;
int n, l[N], r[N], q[N];
modint f[N][N], ans;

inline modint calc(int k) {
	f[0][0] = 1;
	for (int i = 1, t = 0; i <= n << 1; t += q[i] < n, i++)
		for (int j = 0; j <= min(min(i, k), t + (q[i] < n)); j++)
			if (q[i] < n) f[i][j] = f[i-1][j] * (r[q[i]] + 1 - (n + t + k - j)) + (j ? f[i-1][j-1] * (l[q[i]] - (i - 1 - (t - (j - 1)))) : 0);
			else f[i][j] = f[i-1][j] * (r[q[i]] + 1 - (i - 1 - (t - j)));
	return f[n<<1][k];
}

int main() {
	rd(n, P);
	for (int i = 0; i < n << 1; i++)
		l[i] = max(0, (int)ceil(sqrt(n * n - i * i))),
		r[i] = min(2 * n - 1, (int)floor(sqrt(4 * n * n - i * i))),
		q[i+1] = i;
	sort(q + 1, q + (n << 1 | 1), [&](int i, int j) {
		pi x = i < n ? mp(l[i] - 1, r[i]) : mp(r[i], l[i] - 1);
		pi y = j < n ? mp(l[j] - 1, r[j]) : mp(r[j], l[j] - 1);
		return x < y;
	});
	for (int i = 0; i <= n; i++)
		if (i & 1) ans -= calc(i);
		else ans += calc(i);
	print(ans);
	return 0;
}