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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool mapa[10010][10010];
int pot[31], T;
vector<int> tamanhos;
long long K;
int disponivel = 3;
void solve() {
  if (tamanhos[0] == 0) {
    mapa[1][2] = mapa[2][1] = true;
    return;
  }
  int nivel = 1;
  while (nivel <= tamanhos[0]) {
    if (nivel == 1) {
      disponivel += 2;
      mapa[1][3] = mapa[3][1] = mapa[1][4] = mapa[4][1] = true;
    } else {
      disponivel += 2;
      int ant = 2 * nivel;
      int atual = ant + 2;
      mapa[ant][atual] = mapa[atual][ant] = true;
      mapa[ant][atual - 1] = mapa[atual - 1][ant] = true;
      mapa[ant - 1][atual] = mapa[atual][ant - 1] = true;
      mapa[ant - 1][atual - 1] = mapa[atual - 1][ant - 1] = true;
    }
    nivel++;
  }
  mapa[disponivel - 1][2] = mapa[2][disponivel - 1] = mapa[disponivel - 2][2] =
      mapa[2][disponivel - 2] = true;
  for (int i = 1; i < (int)tamanhos.size(); i++) {
    int atual = 2 * (tamanhos[i] + 1);
    if (tamanhos[i] == 0) {
      mapa[1][disponivel] = mapa[disponivel][1] = true;
    } else {
      mapa[atual][disponivel] = mapa[disponivel][atual] = true;
      mapa[atual - 1][disponivel] = mapa[disponivel][atual - 1] = true;
    }
    disponivel++;
    while (++tamanhos[i] < tamanhos[0]) {
      mapa[disponivel][disponivel - 1] = mapa[disponivel - 1][disponivel] =
          true;
      disponivel++;
    }
    mapa[disponivel - 1][2] = mapa[2][disponivel - 1] = true;
  }
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> K;
  pot[0] = 1;
  for (int i = 1; i < 31; i++) pot[i] = 2 * pot[i - 1];
  int i = 31;
  while (K) {
    i--;
    if (K < pot[i]) continue;
    K -= pot[i];
    tamanhos.push_back(i);
  }
  solve();
  cout << disponivel << endl;
  for (int i = 1; i <= disponivel; i++)
    for (int j = 1; j <= disponivel; j++)
      cout << (mapa[i][j] ? "Y" : "N") << (j == disponivel ? "\n" : "");
  return 0;
}