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 PB push_back
#define int long long
#define st first
#define nd second
#define rd third
#define FOR(i, a, b) for(int i =(a); i <=(b); ++i)
#define RE(i, n) FOR(i, 1, n)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define REP(i, n) for(int i = 0;i <(n); ++i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define __builtin_ctz __builtin_ctzll
#define __builtin_clz __builtin_clzll
#define __builtin_popcount __builtin_popcountll
using namespace std;
template<typename TH> void _dbg(const char* sdbg, TH h) { cerr<<sdbg<<"="<<h<<"\n"; }
template<typename TH, typename... TA> void _dbg(const char* sdbg, TH h, TA... t) {
  while(*sdbg != ',') { cerr<<*sdbg++; } cerr<<"="<<h<<","; _dbg(sdbg+1, t...);
}
#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#define debugv(x) {{cerr <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<"\n"; }}
#else
#define debug(...) (__VA_ARGS__)
#define debugv(x)
#define cerr if(0)cout
#endif
#define next ____next
#define prev ____prev
#define left ____left
#define hash ____hash
typedef long long ll;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<pair<int, int> > VPII;
typedef vector<pair<ll, ll> > VPLL;

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
template<class T1, class T2>
ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class A, class B, class C> struct Triple { A first; B second; C third;
  bool operator<(const Triple& t) const { if (st != t.st) return st < t.st; if (nd != t.nd) return nd < t.nd; return rd < t.rd; } };
template<class T> void ResizeVec(T&, vector<int>) {}
template<class T> void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]); sz.erase(sz.begin()); if (sz.empty()) { return; }
  for (T& v : vec) { ResizeVec(v, sz); }
}
typedef Triple<int, int, int> TIII;
template<class A, class B, class C>
ostream& operator<< (ostream &out, Triple<A, B, C> t) { return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")"; }
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class T> ostream& operator<<(ostream& out, set<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class L, class R> ostream& operator<<(ostream& out, map<L, R> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }

VPII Go(int bit) {
  VPII res;
  //res.PB({0, 1});
  RE (i, bit + 1) {
    res.PB({i, 0});
  }
  res.PB({bit + 1, 1});
  res.PB({bit, 1});
  //res.PB({bit, 0});
  FORD (i, bit, 1) {
    res.PB({i, 0});
  }
  return res;
}
void Append(VPII& A, VPII B) {
  A.insert(A.end(), ALL(B));
}
const int N = 1000;
int forb[N];
int need[N];
int32_t main() {

  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  cin.tie(0);
  //double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  

  int n;
  cin>>n;
  REP (i, 1 << n) {
    char c;
    cin>>c;
    if (c == '0') {
      forb[i] = 1;
      need[i] = 1;
    }
  }
  REP (j, 1 << n) {
    REP (i, j) {
      if ((i & j) == i) {
        if (forb[i] == 1 && forb[j] == 0) {
          cout<<"Impossible\n";
          return 0;
        }
        if (forb[i] == 1) {
          need[j] = 0;
        }
      }
    }
  }
  //int any_need = 0;
  PII orig{0, 0};
  VPII pts;
  REP (i, 1 << n) {
    if (need[i] == 1) {
      //any_need = 1;
      VPII cur;
      REP (bit, n) {
        if (i & (1 << bit)) {
          if (cur.empty()) {
            cur = Go(bit);
            continue;
          } 
          VPII rev_cur = cur;
          reverse(ALL(rev_cur));
          VPII heh = Go(bit);
          cur.PB(orig);
          Append(cur, heh);
          cur.PB(orig);
          Append(cur, rev_cur);
          cur.PB(orig);
          reverse(ALL(heh));
          Append(cur, heh);
        }
      }
      pts.PB(orig);
      Append(pts, cur);
    }
  }
  pts.PB(orig);
  cout<<"Possible\n";
  cout<<SZ(pts) - 1<<endl;
  for (auto p : pts) {
    cout<<p.st<<" "<<p.nd<<"\n";
  }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  return 0;
}