Problem: Professor Abacus has just built a new computing engine for making numerical tables. It was designed to calculate the values of a polynomial function in one variable at several points at a time. With the polynomial function f(x) = x2 + 2x + 1, for instance, a possible expected calculation result is 1 (= f(0)), 4 (= f(1)), 9 (= f(2)), 16 (= f(3)), and 25 (= f(4)).

It is a pity, however, the engine seemingly has faulty components and exactly one value among those calculated simultaneously is always wrong. With the same polynomial function as above, it can, for instance, output 1, 4, 12, 16, and 25 instead of 1, 4, 9, 16, and 25.

You are requested to help the professor identify the faulty components. As the first step, you should write a program that scans calculation results of the engine and finds the wrong values.



Input

The input is a sequence of datasets, each representing a calculation result in the following format.

d
v0
v1
...
vd+2


Here, d in the first line is a positive integer that represents the degree of the polynomial, namely, the highest exponent of the variable. For instance, the degree of 4x5 + 3x + 0.5 is five and that of 2.4x + 3.8 is one. d is at most five.

The following d + 3 lines contain the calculation result of f(0), f(1), ... , and f(d + 2) in this order, where f is the polynomial function. Each of the lines contains a decimal fraction between -100.0 and 100.0, exclusive.

You can assume that the wrong value, which is exactly one of f(0), f(1), ... , and f(d+2), has an error greater than 1.0. Since rounding errors are inevitable, the other values may also have errors but they are small and never exceed 10-6.

The end of the input is indicated by a line containing a zero.

Output

For each dataset, output i in a line when vi is wrong.

Example

Input

2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0


Output

2
1
1
6

Solution (in CPP):

Solution:
#define  _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#pragma comment (linker, "/STACK:526000000")

#include "bits/stdc++.h"

using namespace std;
typedef string::const_iterator State;
#define eps 1e-5L
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL
#define seg_size 262144LL
#define MOD 998244353LL
#define pb push_back
#define mp make_pair
typedef long long ll;
#define REP(a,b) for(long long (a) = 0;(a) < (b);++(a))
#define ALL(x) (x).begin(),(x).end()

// geometry library

typedef complex<long double> Point;
typedef pair<complex<long double>, complex<long double>> Line;

typedef struct Circle {
    complex<long double> center;
    long double r;
}Circle;

//内積、 dot(a,b) = |a||b|cos()
long double dot(Point a, Point b) {
    return (a.real() * b.real() + a.imag() * b.imag());
}
//外積、cross(a,b) = |a||b|sin()
long double cross(Point a, Point b) {
    return (a.real() * b.imag() - a.imag() * b.real());
}

//線分と点の距離
long double Dist_Line_Point(Line a, Point b) {
    if (dot(a.second - a.first, b - a.first) < eps) return abs(b - a.first);
    if (dot(a.first - a.second, b - a.second) < eps) return abs(b - a.second);
    return abs(cross(a.second - a.first, b - a.first)) / abs(a.second - a.first);
}

//線分の交差判定
int is_intersected_ls(Line a, Line b) {
    return (cross(a.second - a.first, b.first - a.first) * cross(a.second - a.first, b.second - a.first) < 0) &&
        (cross(b.second - b.first, a.first - b.first) * cross(b.second - b.first, a.second - b.first) < 0);
}

//線分の交点
Point intersection_l(Line a, Line b) {
    Point da = a.second - a.first;
    Point db = b.second - b.first;
    return a.first + da * cross(db, b.first - a.first) / cross(db, da);
}

//線分と線分の距離
long double Dist_Line_Line(Line a, Line b) {
    if (is_intersected_ls(a, b) == 1) {
        return 0;
    }
    return min({ Dist_Line_Point(a,b.first), Dist_Line_Point(a,b.second),Dist_Line_Point(b,a.first),Dist_Line_Point(b,a.second) });
}

//円と円の交点
pair<Point, Point> intersection_Circle_Circle(Circle a, Circle b) {
    long double dist = abs(a.center - b.center);
    assert(dist <= eps + a.r + b.r);
    assert(dist+eps >= abs(a.r - b.r));
    Point target = b.center - a.center;
    long double pointer = target.real() * target.real() + target.imag() * target.imag();
    long double aa = pointer + a.r * a.r - b.r * b.r;
    aa /= 2.0L;
    Point l{ (aa * target.real() + target.imag() * sqrt(pointer * a.r * a.r - aa * aa))/pointer,
            (aa* target.imag() - target.real() * sqrt(pointer * a.r * a.r - aa * aa)) / pointer};
    Point r{ (aa * target.real() - target.imag() * sqrt(pointer * a.r * a.r - aa * aa)) / pointer,
        (aa * target.imag() + target.real() * sqrt(pointer * a.r * a.r - aa * aa)) / pointer };
    r = r + a.center;
    l = l + a.center;
    return mp(l, r);
}

//end of geometry



void init() {
    iostream::sync_with_stdio(false);
    cout << fixed << setprecision(20);
}

unsigned long xor128() {
    static unsigned long x = 123456789, y = 362436069, z = 521288629, w = 88675123;
    unsigned long t = (x ^ (x << 11));
    x = y; y = z; z = w;
    return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}

#define int ll

template<typename A>
vector<A> solve_algebra_equations(vector<vector<A>> now) {
    for (int q = 0; q < now.size(); ++q) {
        if (now[q][q] == 0) {
            for (int j = q + 1; j < now.size(); ++j) {
                if (now[j][q] != 0) {
                    swap(now[j], now[q]);
                    break;
                }
            }
        }
        for (int j = now[q].size() - 1; j >= 0; --j) {
            now[q][j] /= now[q][q];
        }
        for (int j = 0; j < now.size();++j) {
            if (j == q) continue;
            A val = now[j][q];
            for (int t = 0; t < now[j].size(); ++t) {
                now[j][t] -= val * now[q][t];
            }
        }
    }
    vector<A> ans;
    REP(i, now.size()) {
        ans.push_back(now[i].back());
    }
    return ans;
}

void solve(){
    while (true) {
        int d;
        cin >> d;
        if (d == 0) return;
        vector<long double> inputs;
        REP(i, d + 3) {
            long double a;
            cin >> a;
            inputs.push_back(a);
        }
        int ans = -1;
        REP(i, inputs.size()) {
            for (int q = i + 1; q < inputs.size(); ++q) {
                vector<vector<long double>> neko;
                vector<vector<long double>> check;
                REP(j, inputs.size()) {
                    vector<long double> now;
                    now.push_back(inputs[j]);
                    long double geko = 1;
                    for (int t = 0; t <= d; ++t) {
                        now.push_back(geko);
                        geko *= j;
                    }
                    reverse(ALL(now));
                    if (i == j || q == j) {
                        check.push_back(now);
                    }
                    else {
                        neko.push_back(now);
                    }
                 }
                vector<long double>  geko = solve_algebra_equations<long double>(neko);
                REP(t, 2) {
                    long double now_check = 0;
                    for (int q = 0; q < geko.size(); ++q) {
                        now_check += geko[q] * check[t][q];
                    }
                    if (abs(now_check - check[t].back()) < eps) {
                        if (t == 0) {
                            ans = q;
                        }
                        else {
                            ans = i;
                        }
                        break;
                    }
                }
            }
            if (ans != -1) break;
        }
        cout << ans << endl;
    }
}

#undef int
int main() {
    init();
    solve();
}