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;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

const double EPS = 1e-8;
using vec = vector<double>;
using mat = vector<vec>;

// 連立方程式Ax=bを解く
// Aは正方行列を仮定
// 解無し、一意に定まらない : 長さ0のvector
vec gauss_jordan(const mat &A, const vec &b){
    int d = A.size();
    mat B(d, vec(d+1));
    rep(i,d){
        rep(j,d) B[i][j] = A[i][j];
        B[i][d] = b[i];
    }

    rep(i,d){
        int pivot = i;
        for(int j=i; j<d; ++j){
            if( abs(B[j][i]) > abs(B[pivot][i]) ) pivot = j;
        }
        swap(B[i], B[pivot]);

        // 解無し、一意に定まらない : 長さ0のvector
        if(abs(B[i][i]) < EPS) return vec();

        // 注目する変数の係数を1にする
        for(int j=i+1; j<=d; ++j) B[i][j] /= B[i][i];
        rep(j,d){
            if(i==j) continue;
            // j番目の式からi番目の変数を除去
            for(int k=i+1; k<=d; ++k) B[j][k] -= B[j][i]*B[i][k];
        }
    }

    vec ret(d);
    rep(i,d) ret[i] = B[i][d];
    return ret;
}

double f(const vec &a, double x){
    double ret = 0;
    rep(i,a.size()) ret += a[i]*pow(x,i);
    return ret;
}

int solve(int d){
    vector<double> v(d+3);
    rep(i,d+3) cin >>v[i];

    // 考慮しない2つを選択
    rep(i,d+3)rep(j,i){
        int row = 0;
        mat A(d+1,vec(d+1));
        vec b(d+1);

        rep(k,d+3){
            if(k==i || k==j) continue;

            rep(l,d+1) A[row][l] = pow(k,l);
            b[row] = v[k];
            ++row;
        }

        vec a = gauss_jordan(A,b);
        // printf(" %d %d\n", i,j);
        // dbg(a);
        if(a.size() > 0){
            int ct = 0;
            for(int idx:{i,j}){
                // printf(" %.10f %.10f  abs %.10f\n", f(a,idx), v[idx], abs(f(a,idx) - v[idx]));
                if( abs(f(a,idx) - v[idx]) > 1e-5 ) ++ct;
            }
            if(ct==1){
                if( abs(f(a,i) - v[i]) > 1e-5 ) return i;
                else return j;
            }
        }
    }
    assert(false);
    return -1;
}

int main(){
    int d;
    while(cin >>d,d) cout << solve(d) << endl;
    return 0;
}