#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
vector<pair<int, int> > g;
int n, a[1001][1001];
stack<int> s;
int nbits = 29;
int maxbit = -1;
void add(int x, int y) { g.emplace_back(x, y); }
vector<int> v;
void f(int x) {
  if (maxbit == -1) {
    v.emplace_back(1);
    for (int i = 0; i < 2 * nbits - 1; ++i) {
      assert(!s.empty());
      int curr = s.top();
      s.pop();
      add(curr, v.back());
      v.emplace_back(curr);
    }
    add(v.back(), 2);
    v.emplace_back(2);
    for (int i = 1; i <= x; ++i) {
      assert(!s.empty());
      int curr = s.top();
      s.pop();
      add(v[2 * i - 2], curr);
      add(v[2 * i], curr);
    }
  } else {
    int temp = 2 * (maxbit - x);
    vector<int> a;
    a.emplace_back(1);
    for (int i = 1; i <= temp - 1; ++i) {
      assert(!s.empty());
      int curr = s.top();
      s.pop();
      add(a.back(), curr);
      a.emplace_back(curr);
    }
    add(a.back(), v[temp]);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int k;
  cin >> k;
  for (int i = 1000; i > 2; --i) s.push(i);
  for (int i = nbits; i >= 0; --i) {
    if ((k & (1 << i))) {
      f(i);
      if (maxbit == -1) {
        maxbit = i;
      }
    }
  }
  for (auto i : g) {
    a[i.first][i.second] = 1;
    a[i.second][i.first] = 1;
  }
  int n = s.top() - 1;
  cout << s.top() - 1 << '\n';
  ;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (a[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << '\n';
    ;
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."

Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?

Input

The first line contains a single integer k (1 ≤ k ≤ 109).

Output

You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.

The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them. 

Examples

Input

2

Output

4
NNYY
NNYY
YYNN
YYNN

Input

9

Output

8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

Input

1

Output

2
NY
YN

Note

In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.

In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
vector<pair<int, int> > g;
int n, a[1001][1001];
stack<int> s;
int nbits = 29;
int maxbit = -1;
void add(int x, int y) { g.emplace_back(x, y); }
vector<int> v;
void f(int x) {
  if (maxbit == -1) {
    v.emplace_back(1);
    for (int i = 0; i < 2 * nbits - 1; ++i) {
      assert(!s.empty());
      int curr = s.top();
      s.pop();
      add(curr, v.back());
      v.emplace_back(curr);
    }
    add(v.back(), 2);
    v.emplace_back(2);
    for (int i = 1; i <= x; ++i) {
      assert(!s.empty());
      int curr = s.top();
      s.pop();
      add(v[2 * i - 2], curr);
      add(v[2 * i], curr);
    }
  } else {
    int temp = 2 * (maxbit - x);
    vector<int> a;
    a.emplace_back(1);
    for (int i = 1; i <= temp - 1; ++i) {
      assert(!s.empty());
      int curr = s.top();
      s.pop();
      add(a.back(), curr);
      a.emplace_back(curr);
    }
    add(a.back(), v[temp]);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int k;
  cin >> k;
  for (int i = 1000; i > 2; --i) s.push(i);
  for (int i = nbits; i >= 0; --i) {
    if ((k & (1 << i))) {
      f(i);
      if (maxbit == -1) {
        maxbit = i;
      }
    }
  }
  for (auto i : g) {
    a[i.first][i.second] = 1;
    a[i.second][i.first] = 1;
  }
  int n = s.top() - 1;
  cout << s.top() - 1 << '\n';
  ;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (a[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << '\n';
    ;
  }
  return 0;
}

```