Problem:
Problem Statement

We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i.

At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:

Choose a (connected) rope with a total length of at least L, then untie one of its knots.

Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

Constraints

2≤N≤10^5

1≤L≤10^9

1≤a_i≤10^9

All input values are integers.

Input

The input is given from Standard Input in the following format:

N L
a_1 a_2 ... a_n

Output

If it is not possible to untie all of the N-1 knots, print Impossible.

If it is possible to untie all of the knots, print Possible, then print another N-1 lines that describe a possible order to untie the knots. The j-th of those N-1 lines should contain the index of the knot that is untied in the j-th operation. Here, the index of the knot connecting piece i and piece i+1 is i.

If there is more than one solution, output any.

Sample Input 1

3 50
30 20 10

Sample Output 1

Possible
2
1

If the knot 1 is untied first, the knot 2 will become impossible to untie.

Sample Input 2

2 21
10 10

Sample Output 2

Impossible

Sample Input 3

5 50
10 20 30 40 50

Sample Output 3

Possible
1
2
3
4

Another example of a possible solution is 3, 4, 1, 2.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>


using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;


#define int ll
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());


typedef pair<int, int> pii;
typedef pair<string, string> pst;
typedef pair<pii, pii> piii;
typedef vector<piii> vpii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<pii> vpi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;
typedef pair<pld, pld> pldd;
typedef vector<pld> vpd;
typedef complex<double> base;


#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define my_abs(a) (((a) >= 0) ? (a) : -(a))
#define mp(a, b) make_pair(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define MODIK(a) (a >= M ? a - M : a)
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cout << (#x) << " = " << x << endl;
#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif


const double PI = 2 * acos(0.0);

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";

istream &operator >> (istream &in, pii &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pii &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


istream &operator >> (istream &in, pld &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pld &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) <<
               ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void print(vi kek, int k)
{
    ofstream cout("output.txt");
    cout << SZ(kek) << ' ' << k << '\n';
    for (auto p : kek) {
        cout << p << ' ';
    }
    cout.close();
    exit(0);
}


inline bool break_time(ld time)
{
    return clock() > CLOCKS_PER_SEC * time;
}


ld get_rand_ld(ld l, ld r)
{
    uniform_real_distribution<double> gen(l, r);
    return gen(rng);
}


void smain();

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

#ifdef ART_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(4) << fixed;
    smain();
#ifdef ART_HOME
    cout << "\n\n\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
}


const int M1 = 1000000093, p1 = 27162;
const int M2 = 1000000087, p2 = 241;


const int N = 1e5 + 100, B = 330, oo = 1e9, A = 26, M = 1e6 + 7, pw = 3;


int cnt[N];


void smain()
{
    int n, l;
    cin >> n >> l;

    vi lel(n);
    for (int i = 0; i < n; ++i)
        cin >> lel[i];

    int label = 0;

    for (int i = 0; i + 1 < n; ++i)
        if (lel[label] + lel[label + 1] < lel[i] + lel[i + 1])
            label = i;


    if (lel[label] + lel[label + 1] < l)
    {
        cout << "Impossible\n";
    }
    else
    {
        cout << "Possible\n";
        for (int i = 0; i < label; ++i)
            cout << i + 1 << '\n';

        for (int i = n - 2; i > label; --i)
            cout << i + 1 << '\n';

        cout << label + 1;
    }
}