Problem: Given are a positive integer N and a sequence of length 2^N consisting of 0s and 1s: A_0,A_1,\ldots,A_{2^N-1}. Determine whether there exists a closed curve C that satisfies the condition below for all 2^N sets S \subseteq \\{0,1,\ldots,N-1 \\}. If the answer is yes, construct one such closed curve.

* Let x = \sum_{i \in S} 2^i and B_S be the set of points \\{ (i+0.5,0.5) | i \in S \\}.
* If there is a way to continuously move the closed curve C without touching B_S so that every point on the closed curve has a negative y-coordinate, A_x = 1.
* If there is no such way, A_x = 0.



For instruction on printing a closed curve, see Output below.

Constraints

* 1 \leq N \leq 8
* A_i = 0,1 \quad (0 \leq i \leq 2^N-1)
* A_0 = 1

Input

Input is given from Standard Input in the following format:


N
A_0A_1 \cdots A_{2^N-1}


Output

If there is no closed curve that satisfies the condition, print `Impossible`.

If such a closed curve exists, print `Possible` in the first line. Then, print one such curve in the following format:


L
x_0 y_0
x_1 y_1
:
x_L y_L


This represents the closed polyline that passes (x_0,y_0),(x_1,y_1),\ldots,(x_L,y_L) in this order.

Here, all of the following must be satisfied:

* 0 \leq x_i \leq N, 0 \leq y_i \leq 1, and x_i, y_i are integers. (0 \leq i \leq L)
* |x_i-x_{i+1}| + |y_i-y_{i+1}| = 1. (0 \leq i \leq L-1)
* (x_0,y_0) = (x_L,y_L).



Additionally, the length of the closed curve L must satisfy 0 \leq L \leq 250000.

It can be proved that, if there is a closed curve that satisfies the condition in Problem Statement, there is also a closed curve that can be expressed in this format.

Examples

Input

1
10


Output

Possible
4
0 0
0 1
1 1
1 0
0 0


Input

2
1000


Output

Possible
6
1 0
2 0
2 1
1 1
0 1
0 0
1 0


Input

2
1001


Output

Impossible


Input

1
11


Output

Possible
0
1 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using pint = pair<int, int>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define REP(i, n) FOR(i,0,n)

const string YES = "Possible";
const string NO = "Impossible";

bool impossible_checker(int N, string S) {
    assert((int)(S.length()) == (1 << N));
    if (S[0] == '0') return true;
    REP(a, 1 << N)
    {
        REP(b, 1 << N) if (((~b) & a) == 0) {
            if (S[b] == '1' and S[a] == '0') return true;
        }
    }
    return false;
}

vector<vector<int>> S(9);

int x_now, y_now;

vector<pint> ret;

void up() {
    y_now++;
    ret.emplace_back(x_now, y_now);
}
void down() {
    y_now--;
    ret.emplace_back(x_now, y_now);
}
void right() {
    x_now++;
    ret.emplace_back(x_now, y_now);
}
void left() {
    x_now--;
    ret.emplace_back(x_now, y_now);
}

void do_rloop(int X)
{
    if (y_now == 0) {
        up();
    }
    while (x_now < X) right();
    while (x_now > X) left();
    right();
    down();
    left();
    up();
}

void do_lloop(int X) {
    if (y_now == 0) {
        up();
    }
    while (x_now < X) right();
    while (x_now > X) left();
    down();
    right();
    up();
    left();
}

void go_origin() {
    if (y_now == 0) up();
    while (x_now > 0) left();
    down();
}

int main()
{
    int N;
    string A;
    cin >> N >> A;
    if (impossible_checker(N, A)) {
        cout << NO << endl;
        return 0;
    }

    S[1] = {+1};
    S[2] = {+1, +2, -1, -2};
    FOR(d, 3, S.size()) {
        S[d] = {d};
        vector<int> prev = S[d - 1];
        S[d].insert(S[d].end(), ALL(prev));
        S[d].push_back(-d);
        reverse(ALL(prev));
        for (auto &x : prev) x *= -1;
        S[d].insert(S[d].end(), ALL(prev));
    }

    up();
    down();

    REP(subset, 1 << N) if (A[subset] == '0') {
        bool must = true;
        REP(ss, 1 << N) if (((~subset) & ss) == 0 and ss < subset) {
            if (A[ss] == '0') must = false;
        }
        if (!must) continue;
        int cnt = __builtin_popcount(subset);
        if (cnt == 1) {
            int i = 0;
            while ((subset >> i) % 2 == 0) i++;
            do_rloop(i);
            continue;
        }

        vector<int> ns;
        REP(i, N) if ((subset >> i) & 1) ns.push_back(i);
        for (auto dict : S[cnt]) {
            int id_ = abs(dict) - 1;
            if (dict > 0) do_rloop(ns[id_]);
            else do_lloop(ns[id_]);
        }
    }
    go_origin();
    up();

    cout << YES << "\n";
    cout << ret.size() - 1 << "\n";
    for (auto p : ret) cout << p.first << " " << p.second << "\n";
}