#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1010;
const int START = 0;
const int FINISH = 1;
bool g[MAXN][MAXN];
int n, paths;
void add_edge(int a, int b) { g[a][b] = g[b][a] = true; }
void add_pow2(int k, int len) {
  if (k == len) {
    for (int i = 0; i < k; ++i) {
      if (i == 0) {
        add_edge(n++, START);
        add_edge(n++, START);
      } else {
        add_edge(n, n - 1);
        add_edge(n, n - 2);
        add_edge(n + 1, n - 1);
        add_edge(n + 1, n - 2);
        n += 2;
      }
      if (i == k - 1) {
        add_edge(n - 1, FINISH);
        add_edge(n - 2, FINISH);
      }
    }
  }
  if (k < len) {
    if (k == 0) {
      add_edge(n++, START);
    } else {
      add_edge(2 * k, n);
      add_edge(2 * k + 1, n);
      n++;
    }
    for (int i = k + 1; i < len; ++i) {
      add_edge(n, n - 1);
      n++;
    }
    add_edge(n - 1, FINISH);
  }
}
int main() {
  cin >> paths;
  if (paths == 1) {
    cout << "2\nNY\nYN\n";
    return 0;
  }
  n = 2;
  int len = -1;
  for (int k = 30; k >= 0; --k) {
    if (paths & (1 << k)) {
      if (len == -1) len = k;
      add_pow2(k, len);
    }
  }
  cout << n << endl;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j)
      if (g[i][j])
        cout << "Y";
      else
        cout << "N";
    cout << endl;
  }
  return 0;
}

### Prompt

Please formulate 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 int MAXN = 1010;
const int START = 0;
const int FINISH = 1;
bool g[MAXN][MAXN];
int n, paths;
void add_edge(int a, int b) { g[a][b] = g[b][a] = true; }
void add_pow2(int k, int len) {
  if (k == len) {
    for (int i = 0; i < k; ++i) {
      if (i == 0) {
        add_edge(n++, START);
        add_edge(n++, START);
      } else {
        add_edge(n, n - 1);
        add_edge(n, n - 2);
        add_edge(n + 1, n - 1);
        add_edge(n + 1, n - 2);
        n += 2;
      }
      if (i == k - 1) {
        add_edge(n - 1, FINISH);
        add_edge(n - 2, FINISH);
      }
    }
  }
  if (k < len) {
    if (k == 0) {
      add_edge(n++, START);
    } else {
      add_edge(2 * k, n);
      add_edge(2 * k + 1, n);
      n++;
    }
    for (int i = k + 1; i < len; ++i) {
      add_edge(n, n - 1);
      n++;
    }
    add_edge(n - 1, FINISH);
  }
}
int main() {
  cin >> paths;
  if (paths == 1) {
    cout << "2\nNY\nYN\n";
    return 0;
  }
  n = 2;
  int len = -1;
  for (int k = 30; k >= 0; --k) {
    if (paths & (1 << k)) {
      if (len == -1) len = k;
      add_pow2(k, len);
    }
  }
  cout << n << endl;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j)
      if (g[i][j])
        cout << "Y";
      else
        cout << "N";
    cout << endl;
  }
  return 0;
}

```