Problem: Wind Corridor is a covered passageway where strong wind is always blowing. It is a long corridor of width W, and there are several pillars in it. Each pillar is a right prism and its face is a polygon (not necessarily convex).

In this problem, we consider two-dimensional space where the positive x-axis points the east and the positive y-axis points the north. The passageway spans from the south to the north, and its length is infinity. Specifically, it covers the area 0 ≤ x ≤ W. The outside of the passageway is filled with walls. Each pillar is expressed as a polygon, and all the pillars are located within the corridor without conflicting or touching each other.

Wind blows from the south side of the corridor to the north. For each second, w unit volume of air can be flowed at most if the minimum width of the path of the wind is w. Note that the path may fork and merge, but never overlaps with pillars and walls.

Your task in this problem is to write a program that calculates the maximum amount of air that can be flowed through the corridor per second.



Input

The input consists of multiple datasets. Each dataset has the following format:

The first line of the input contains two integers W and N. W is the width of the corridor, and N is the number of pillars. W and N satisfy the following condition: 1 ≤ W ≤ 104 and 0 ≤ N ≤ 200.

Then, N specifications of each pillar follow. Each specification starts with a line that contains a single integer M, which is the number of the vertices of a polygon (3 ≤ M ≤ 40). The following M lines describe the shape of the polygon. The i-th line (1 ≤ i ≤ M) contains two integers xi and yi that denote the coordinate of the i-th vertex (0 < xi < W, 0 < yi < 104).

The last dataset is followed by a line containing two zeros. This line is not a part of any dataset and should not be processed.

Output

For each dataset, your program should print a line that contains the maximum amount of air flow per second, in unit volume. The output may contain arbitrary number of digits after the decimal point, but the absolute error must not exceed 10-6.

Example

Input

5 2
4
1 1
1 2
2 2
2 1
4
3 3
3 4
4 4
4 3
0 0


Output

3.41421356

Solution (in CPP):

Solution:
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

using ld = double;
using Point =  complex<ld>;
const ld eps = 1e-9;
const ld pi = acos(-1.0);

bool eq(ld a, ld b) {
	return (abs(a - b) < eps);
}

bool cmp(Point x,Point y){
	if(eq(x.real(),y.real()))return x.imag()<y.imag();
	return x.real()<y.real();
}

bool eqq(Point x,Point y){
	return eq(x.real(),y.real())&&eq(x.imag(),y.imag());
}
//内積
ld dot(Point a, Point b) {
	return real(conj(a)*b);
}
//外積
ld cross(Point a, Point b) {
	return imag(conj(a)*b);
}


//線分
//直線にするなら十分通い２点を端点とすればよい
class Line {
public:
	Point a, b;
};
//円
class Circle {
public:
	Point p;
	ld r;
};
//3点の位置関係
int ccw(Point a, Point b, Point c) {
	b -= a; c -= a;
	if (cross(b, c) > eps)return 1;//a,b,cが反時計回り
	if (cross(b, c) < -eps)return -1;//a,b,cが時計回り
	if (dot(b, c) < 0)return 2;//c,a,bの順に一直線
	if (norm(b) < norm(c))return -2;//a,b,cの順に一直線
	return 0;//a,c,bの順に一直線
}
//2直線の交差判定
bool isis_ll(Line l, Line m) {
	return !eq(cross(l.b - l.a, m.b - m.a), 0);
}
//直線と線分の交差判定
bool isis_ls(Line l, Line s) {
	return (cross(l.b - l.a, s.a - l.a)*cross(l.b - l.a, s.b - l.a) < eps);
}
//点が直線上に存在するか
bool isis_lp(Line l, Point p) {
	return (abs(cross(l.b - p, l.a - p)) < eps);
}
//点が線分上に存在するか
bool isis_sp(Line s, Point p) {
	return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}
//線分と線分の交差判定
bool isis_ss(Line s, Line t) {
	if (isis_sp(s, t.a) || isis_sp(s, t.b) || isis_sp(t, s.a) || isis_sp(t, s.b))return true;
	return(cross(s.b - s.a, t.a - s.a)*cross(s.b - s.a, t.b - s.a) < -eps && cross(t.b - t.a, s.a - t.a)*cross(t.b - t.a, s.b - t.a) < -eps);
}
//点から直線への垂線の足
Point proj(Line l, Point p) {
	ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}

Point cont(Line l, Point p){
	return 2.0*proj(l,p) - p;
}
//直線と直線の交点
//平行な２直線に対しては使うな！！！！
Point is_ll(Line s, Line t) {
	Point sv = s.b - s.a; Point tv = t.b - t.a;
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}
//直線と点の距離
ld dist_lp(Line l, Point p) {
	return abs(p - proj(l, p));
}
//直線と直線の距離
ld dist_ll(Line l, Line m) {
	return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}
//線分と直線の距離
ld dist_ls(Line l, Line s) {
	return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}
//線分と点の距離
ld dist_sp(Line s, Point p) {
	Point r = proj(s, p);
	return isis_sp(s, r) ? abs(p - r) : min(abs(p - s.a), abs(p - s.b));
}
//線分と線分の距離
ld dist_ss(Line s, Line t) {
	if (isis_ss(s, t))return 0;
	return min({ dist_sp(s,t.a),dist_sp(s,t.b),dist_sp(t,s.a),dist_sp(t,s.b) });
}


void solve(int w,int n){
    vector<vector<Point>> pols(n);
    rep(i,n){
        int m;
        cin>>m;
        rep(j,m){
            int x,y;
            cin>>x>>y;
            pols[i].push_back(Point(x,y));
        }
    }
    ld dist[n+2][n+2];
    rep(i,n+2)dist[i][i]=0;
    rep(i,n)rep(j,i){
        ld mi=inf;
        for(auto p : pols[i]){
            rep(k,pols[j].size()){
                mi=min(mi,dist_sp(Line{pols[j][k],pols[j][(k+1)%pols[j].size()]},p));
            }
        }
        for(auto p : pols[j]){
            rep(k,pols[i].size()){
                mi=min(mi,dist_sp(Line{pols[i][k],pols[i][(k+1)%pols[i].size()]},p));
            }
        }
        dist[i][j]=dist[j][i]=mi;
    }
    rep(i,n){
        ld mi=inf,ma=0;
        for(auto p : pols[i]){
            mi=min(mi,p.real());
            ma=max(ma,p.real());
        }
        dist[i][n]=dist[n][i]=mi;
        dist[i][n+1]=dist[n+1][i]=(ld)w-ma;
    }
    dist[n][n+1]=dist[n+1][n]=(ld)w;
    rep(i,n+2)rep(j,n+2)rep(k,n+2){
        dist[j][k]=min(dist[j][k],dist[j][i]+dist[i][k]);
    }
    cout<<dist[n][n+1]<<endl;
}
int main(){
    int w,n;
    cout<<fixed<<setprecision(12);
    while(cin>>w>>n,w!=0)solve(w,n);
    return 0;
}