#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <ctime>
#include <algorithm>
#include <sstream>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <iterator>
#include <functional>
#include <bitset>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
clock_t __my_start_clock = clock();
#else
#define eprintf(...)
#endif

#define TIMESTAMP(x) eprintf("["#x"] Time : %.3lf s.\n", (clock() - __my_start_clock)*1.0/CLOCKS_PER_SEC)
#define TIMESTAMPf(x,...) eprintf("[" x "] Time : %.3lf s.\n", __VA_ARGS__, (clock() - __my_start_clock)*1.0/CLOCKS_PER_SEC)

using namespace std;

#ifdef LOCAL
static struct __timestamper {
  string what;
  __timestamper(const char* what) : what(what){};
  __timestamper(const string& what) : what(what){};
  ~__timestamper(){
    TIMESTAMPf("%s", what.data());
	}
} __TIMESTAMPER("end");
#else 
struct __timestamper {};
#endif

typedef long long ll;
typedef long double ld;


bool solve(int n, string s) {
  if (s[0] == '0')
    return false;
  vector<int> bads;
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if (!(i & (1 << j))) {
        continue;
      }
    }
    bool found_zero_sub = false;
    for (int j = 0; j < n; j++) {
      if (!(i & (1 << j))) {
        continue;
      }
      if (s[i ^ (1 << j)] == '0') {
        found_zero_sub = true;
      }
    }
    if (found_zero_sub) {
      if (s[i] == '1') {
        return false;
      }
    } else {
      if (s[i] == '0') {
        bads.push_back(i);
      }
    }
  }

  vector<int> seq;

  for (int x : bads) {
    vector<int> ss;
    for (int i = 0; i < n; i++) {
      if (!(x & (1 << i))) {
        continue;
      }
      int cur = ss.size();
      ss.push_back(i + 1);
      for (int j = cur - 1; j >= 0; j--) {
        ss.push_back(-ss[j]);
      }
      if (cur) {
        ss.push_back(-(i + 1));
      }
    }
    seq.insert(seq.end(), ss.begin(), ss.end());
  }

  printf("Possible\n");

  vector<pair<int, int>> pts;

  pts.emplace_back(0, 0);

  auto push = [&] (int x,int y) {
    if (pts.size() > 1 && pts[pts.size() - 2].first == x && pts[pts.size() - 2].second == y) {
      pts.pop_back();
    } else {
      pts.emplace_back(x, y);
    }
  };

  for (int x : seq) {
    if (x < 0) {
      x = -x;
      for (int i = 1; i <= x; i++) {
        push(0, i);
      }
      push(1, x);
      push(1, x - 1);
      for (int i = x - 1; i >= 0; i--) {
        push(0, i);
      }
    } else {
      for (int i = 1; i < x; i++) {
        push(0, i);
      }
      push(1, x - 1);
      push(1, x);
      for (int i = x ; i >= 0; i--) {
        push(0, i);
      }
    }
  }

  printf("%d\n", (int)pts.size() - 1);
  for (auto p : pts) {
    printf("%d %d\n", p.second, p.first);
  }
  return true;
}

int main() {
#ifdef LOCAL
  freopen("e.in", "r", stdin);
  freopen("e.out", "w", stdout);
#endif

  int n;
  while (scanf("%d", &n) == 1) {
    char s[512];
    scanf("%s", s);
    if (!solve(n, s)) {
      printf("Impossible\n");
    }
  }

  return 0;
}

### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <ctime>
#include <algorithm>
#include <sstream>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <iterator>
#include <functional>
#include <bitset>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
clock_t __my_start_clock = clock();
#else
#define eprintf(...)
#endif

#define TIMESTAMP(x) eprintf("["#x"] Time : %.3lf s.\n", (clock() - __my_start_clock)*1.0/CLOCKS_PER_SEC)
#define TIMESTAMPf(x,...) eprintf("[" x "] Time : %.3lf s.\n", __VA_ARGS__, (clock() - __my_start_clock)*1.0/CLOCKS_PER_SEC)

using namespace std;

#ifdef LOCAL
static struct __timestamper {
  string what;
  __timestamper(const char* what) : what(what){};
  __timestamper(const string& what) : what(what){};
  ~__timestamper(){
    TIMESTAMPf("%s", what.data());
	}
} __TIMESTAMPER("end");
#else 
struct __timestamper {};
#endif

typedef long long ll;
typedef long double ld;


bool solve(int n, string s) {
  if (s[0] == '0')
    return false;
  vector<int> bads;
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if (!(i & (1 << j))) {
        continue;
      }
    }
    bool found_zero_sub = false;
    for (int j = 0; j < n; j++) {
      if (!(i & (1 << j))) {
        continue;
      }
      if (s[i ^ (1 << j)] == '0') {
        found_zero_sub = true;
      }
    }
    if (found_zero_sub) {
      if (s[i] == '1') {
        return false;
      }
    } else {
      if (s[i] == '0') {
        bads.push_back(i);
      }
    }
  }

  vector<int> seq;

  for (int x : bads) {
    vector<int> ss;
    for (int i = 0; i < n; i++) {
      if (!(x & (1 << i))) {
        continue;
      }
      int cur = ss.size();
      ss.push_back(i + 1);
      for (int j = cur - 1; j >= 0; j--) {
        ss.push_back(-ss[j]);
      }
      if (cur) {
        ss.push_back(-(i + 1));
      }
    }
    seq.insert(seq.end(), ss.begin(), ss.end());
  }

  printf("Possible\n");

  vector<pair<int, int>> pts;

  pts.emplace_back(0, 0);

  auto push = [&] (int x,int y) {
    if (pts.size() > 1 && pts[pts.size() - 2].first == x && pts[pts.size() - 2].second == y) {
      pts.pop_back();
    } else {
      pts.emplace_back(x, y);
    }
  };

  for (int x : seq) {
    if (x < 0) {
      x = -x;
      for (int i = 1; i <= x; i++) {
        push(0, i);
      }
      push(1, x);
      push(1, x - 1);
      for (int i = x - 1; i >= 0; i--) {
        push(0, i);
      }
    } else {
      for (int i = 1; i < x; i++) {
        push(0, i);
      }
      push(1, x - 1);
      push(1, x);
      for (int i = x ; i >= 0; i--) {
        push(0, i);
      }
    }
  }

  printf("%d\n", (int)pts.size() - 1);
  for (auto p : pts) {
    printf("%d %d\n", p.second, p.first);
  }
  return true;
}

int main() {
#ifdef LOCAL
  freopen("e.in", "r", stdin);
  freopen("e.out", "w", stdout);
#endif

  int n;
  while (scanf("%d", &n) == 1) {
    char s[512];
    scanf("%s", s);
    if (!solve(n, s)) {
      printf("Impossible\n");
    }
  }

  return 0;
}

```