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
#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef tuple<int,int,int> T;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
typedef vector<T> vt;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("YES");}else{puts("NO");}

// Mod int
int mod = 998244353;
struct mint {
  ll x;
  mint():x(0){}
  mint(ll x):x((x%mod+mod)%mod){}
  // mint(ll x):x(x){}
  mint& fix() { x = (x%mod+mod)%mod; return *this;}
  mint operator-() const { return mint(0) - *this;}
  mint& operator+=(const mint& a){ if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a){ (x*=a.x)%=mod; return *this;}
  mint operator+(const mint& a)const{ return mint(*this) += a;}
  mint operator-(const mint& a)const{ return mint(*this) -= a;}
  mint operator*(const mint& a)const{ return mint(*this) *= a;}
  bool operator<(const mint& a)const{ return x < a.x;}
  bool operator==(const mint& a)const{ return x == a.x;}
};
istream& operator>>(istream&i,mint&a){i>>a.x;return i;}
ostream& operator<<(ostream&o,const mint&a){o<<a.x;return o;}
typedef vector<mint> vm;
//

const int MX = 255;
mint dp[MX][MX][MX];

int main() {
  int n = 200;
  cin>>n>>mod;
  int n2 = n*2;
  vi l(n), r(n), r2(n);
  rep(i,n2) {
    int nl = INF, nr = -INF;
    rep(j,n2) {
      int x = i*i+j*j;
      if (n*n <= x && x <= n2*n2) {
        mins(nl,j);
        maxs(nr,j);
      }
    }
    ++nr;
    if (i < n) l[i] = nl, r[i] = nr;
    else r2[i-n] = nr;
  }

  reverse(rng(l));
  reverse(rng(r));
  reverse(rng(r2));
  // cerr<<l<<endl;cerr<<r<<endl;cerr<<r2<<endl;


  rep(i,n+1) dp[0][i][0] = 1;
  int ri = 0;
  rep(i,n) {
    int ni = i+1;
    int si = ri, ti = si;
    while (ti < n && r2[ti] < l[i]) ++ti;
    rep(k,n+1) {

      mint x2 = 1;
      srep(j,si,ti) {
        int x = r2[j]-(j+k);
        maxs(x,0);
        x2 *= x;
      }

      srep(j,k,n+1) {
        // mint now = dp[i][j][k]*x2;
        int x1 = r[i] - (n+i+(j-k));
        if (x1 > 0) {
          dp[ni][j][k] += dp[i][j][k]*x1*x2;
        }
        if (k < j) {
          int x = l[i]-(ti+k);
          if (x > 0) dp[ni][j][k+1] += dp[i][j][k]*x2*x;
        }
      }
    }
    ri = ti;
  }

  // rep(i,n+1)rep(j,n+1)rep(k,j+1) cout<<i<<" "<<j<<" "<<k<<": "<<dp[i][j][k]<<endl;

  mint ans = 0;
  rep(i,n+1) {
    int sg = (i&1)?-1:1;
    mint now = dp[n][i][i];
    srep(j,ri,n) {
      int x = r2[j]-(j+i);
      maxs(x,0);
      now *= x;
    }
    // cerr<<now<<endl;
    ans += now*sg;
  }
  cout<<ans<<endl;
  return 0;
}




















