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;
template <typename t>
inline t getnum() {
  t res = 0;
  char c;
  int sign = 1;
  while ((c = getchar()) == ' ' || c == '\n') {
    c = getchar();
    if (c == ' ' || c == '\n')
      continue;
    else
      break;
  }
  if (c == '+') {
  } else if (c == '-')
    sign = -1;
  else
    res = c - '0';
  while (1) {
    c = getchar();
    if (c >= '0' && c <= '9')
      res = res * 10 + c - '0';
    else
      break;
  }
  return res * sign;
}
inline string getstr() {
  string str;
  char k;
  while ((k = getchar()) == ' ' || k == '\n') {
    k = getchar();
    if (k == ' ' || k == '\n')
      continue;
    else
      break;
  }
  while ((k = getchar()) != EOF && k != '\n' && k != '\t' && k != '\v' &&
         k != '\0' && k != ' ')
    str.push_back(k);
  return str;
}
int nd = 2, adj[1000][1000];
int main() {
  int k = getnum<int>();
  vector<bool> bit;
  memset(adj, 0, sizeof(adj));
  while (k) {
    bit.push_back(k % 2);
    k /= 2;
  }
  for (int i = 0; i < bit.size() - 1; i++) {
    adj[nd - 1][nd + 1] = adj[nd + 1][nd - 1] = 1;
    adj[nd - 1][nd + 2] = adj[nd + 2][nd - 1] = 1;
    if (nd > 2) {
      adj[nd][nd + 1] = adj[nd + 1][nd] = 1;
      adj[nd][nd + 2] = adj[nd + 2][nd] = 1;
    }
    nd += 2;
  }
  if (nd > 2) {
    adj[nd - 1][2] = adj[2][nd - 1] = 1;
    adj[nd][2] = adj[2][nd] = 1;
  }
  for (int i = bit.size() - 2; i >= 1; i--)
    if (bit[i]) {
      for (int j = 0; j < bit.size() - i - 1; j++) {
        nd++;
        adj[(j ? nd - 1 : 1)][nd] = adj[nd][(j ? nd - 1 : 1)] = 1;
      }
      adj[nd][2 * (bit.size() - i) + 1] = adj[2 * (bit.size() - i) + 1][nd] = 1;
      adj[nd][2 * (bit.size() - i) + 2] = adj[2 * (bit.size() - i) + 2][nd] = 1;
    }
  if (bit[0]) {
    int init = 1;
    for (int i = 0; i < bit.size() - 1; i++) {
      nd++;
      adj[init][nd] = adj[nd][init] = 1;
      init = nd;
    }
    if (init - 2) adj[init][2] = adj[2][init] = 1;
  }
  cout << nd;
  puts("");
  for (int i = 1; i <= nd; i++) {
    for (int j = 1; j <= nd; j++) cout << (adj[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}