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>

#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

using namespace std;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }

vi sift(vi a, int x) {
    vi b;
    for (int y: a) if (abs(y) != x) b.pb(y);
    return b;
}

bool trivial(vi a) {
    vi st;
    for (int x: a) {
        if (st.empty() || st.back() + x) st.pb(x);
        else st.pop_back();
    }
    return st.empty();
}

int n, l;

bool good(vi a) {
    if (trivial(a)) return false;
    for1(i, n) if (!trivial(sift(a, i))) return false;
    return true;
}

vi best;

void rec(vi &a) {
    if (a.size() >= best.size()) return;
    if (good(a)) {
        best = a;
        return;
    }
    for1(i, n) {
        a.pb(i);
        rec(a);
        a.pop_back();

        a.pb(-i);
        rec(a);
        a.pop_back();
    }
}

void act(vi v, vector<pii> &p) {
    for (int x: v) {
        for1(i, abs(x) - 1) p.pb({i, 0});
        if (x > 0) {
            p.pb({x, 0});
            p.pb({x, 1});
            p.pb({x - 1, 1});
            p.pb({x - 1, 0});
        } else {
            p.pb({-x - 1, 1});
            p.pb({-x, 1});
            p.pb({-x, 0});
            p.pb({-x - 1, 0});
        }
        ford(i, abs(x) - 1) p.pb({i, 0});
    }
}

vi seq(vi c) {
    if (c.size() == 1) return c;
    int x = c.back();
    c.pop_back();
    vi res = seq(c);
    vi ans = res;
    ans.pb(x);
    reverse(all(res));
    for (int y: res) ans.pb(-y);
    ans.pb(-x);
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(10);
    cout << fixed;
#ifdef LOCAL_DEFINE
    freopen("input.txt", "rt", stdin);
#endif

    int n;
    string s;
    cin >> n >> s;
    vi good(1 << n);
    forn(i, 1 << n) good[i] = s[i] - '0';

    bool ok = good[0];
    forn(i, 1 << n) forn(j, n) {
        if ((i >> j) & 1) continue;
        if (!good[i] && good[i | (1 << j)]) ok = false;
    }

    if (!ok) cout << "Impossible\n";
    else {
        cout << "Possible\n";
        vector<pii> ans = {{0, 0}};
        forn(i, 1 << n) {
            if (good[i]) continue;
            bool ok = true;
            forn(j, n) if ((i >> j) & 1) ok &= good[i ^ (1 << j)];
            if (!ok) continue;
            vi c;
            forn(j, n) if ((i >> j) & 1) c.pb(j + 1);
            act(seq(c), ans);
        }
        cout << ans.size() - 1 << '\n';
        for (auto w: ans) cout << w.fi << ' ' << w.se << '\n';
    }

#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}