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:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using u64 = uint_fast64_t; 
using pii =  pair<int, int>;
using pll =  pair<long long, long long>;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
constexpr char ln =  '\n';
////////////////////////////////////////////////////////////////////////////////

using DD = double;
constexpr DD EPS = 1e-3;
inline int sgn(DD a) {return (a < -EPS) ? -1 : (a > EPS) ? 1 : 0;}

template<class T>
struct Matrix {

    int N,M;
    vector<vector<T>> A;

    Matrix() {}
    Matrix(int N, int M, T x = 0) : N(N), M(M), A(N, vector<T>(M, x)) {}

    void init(int n, int m, T x = 0) {A.assign(n, vector<T>(m, x));}
    int size() const {return N;}
    inline vector<T>& operator [] (int i) {return A[i];}

    int GaussJordan(bool is_extended = false) {
        int rank = 0;
        for (int col = 0; col < M; ++col) {
            // 拡大係数行列の場合は最後の列は掃き出ししない
            if (is_extended && col == M-1) break;

            // ピボットを探す
            int pivot = -1;
            T ma = EPS;
            for (int row = rank; row < N; ++row) {
                if (abs(A[row][col]) > ma) {
                    ma = abs(A[row][col]);
                    pivot = row;
                }
            }
            // ピボットがなかったら次の列へ
            if (pivot == -1) continue;

            // まずは行を swap
            swap(A[pivot], A[rank]);

            // ピボットの値を 1 にする
            auto fac = A[rank][col];
            for (int col2 = 0; col2 < M; ++col2) A[rank][col2] /= fac;

            // ピボットのある列の値がすべて 0 になるように掃き出す
            for (int row = 0; row < N; ++row) {
                if (row != rank && sgn(A[row][col])) {
                    auto fac = A[row][col];
                    for (int col2 = 0; col2 < M; ++col2) {
                        A[row][col2] -= A[rank][col2] * fac;
                    }
                }
            }
            ++rank;
        }
        return rank;
    }

    vector<T> linear_equation(const vector<T> &b) {
        // extended
        Matrix<T> C(N, M + 1);
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) C[i][j] = A[i][j];
            C[i][M] = b[i];
        }
        int rank = C.GaussJordan(true);
        
        // check if it has no solution
        vector<T> res;
        for (int row = rank; row < N; ++row) if (sgn(C[row][M])) return res;

        // answer
        res.assign(M, 0);
        for (int i = 0; i < rank; ++i) res[i] = C[i][M];
        return res;
    }
};

////////////////////////////////////////////////////////////////////////////////

void AOJ1328() {
    auto solve=[](int N) {
        vector<DD> d(N+3);
        rep(i,N+3) cin >> d[i];

        vector<vector<DD>> beki(N+3, vector<DD>(N+1,1));
        rep(i,N+3) rep(j,N) beki[i][j+1] = beki[i][j]*i;

        auto calc=[&](const vector<DD> &ans, int x) {
            DD ret = 0;
            DD b = 1;
            rep(i,N+1) {
                ret += ans[i]*b;
                b *= x;
            }
            return ret;
        };
        rep(x,N+3) {
            for (int y = x+1; y < N+3; y++) {
                Matrix<DD> mat(N+1, N+1);
                vector<vector<DD>> A;
                vector<DD> B;
                rep(i,N+3) {
                    if (i==x or i==y) continue;
                    A.emplace_back(beki[i]);
                    B.emplace_back(d[i]);
                }
                rep(i,N+1) rep(j,N+1) mat[i][j] = A[i][j];
                auto ret = mat.linear_equation(B);
                if (!ret.empty()) {
                    DD ansx = calc(ret,x);
                    DD ansy = calc(ret,y);
                    if (!sgn(d[x]-ansx)) {
                        cout << y << ln;
                        return;
                    }
                    if (!sgn(d[y]-ansy)) {
                        cout << x << ln;
                        return;
                    }
                }
            }
        }
    };
    while (1) {
        int n; cin >> n;
        if (!n) break;
        solve(n);
    }
}
int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr);
    AOJ1328();
}