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.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1050;
int res[N][N];
int ptr;
void add_edge(int u, int v) { res[u][v] = res[v][u] = 1; }
void endd() {
  cout << ptr << "\n";
  for (int i = 0; i < ptr; ++i) {
    for (int j = 0; j < ptr; ++j) {
      cout << "NY"[res[i][j]];
    }
    cout << "\n";
  }
  exit(0);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int k;
  cin >> k;
  if (k == 3) {
    add_edge(0, 2);
    add_edge(0, 3);
    add_edge(0, 4);
    add_edge(2, 1);
    add_edge(3, 1);
    add_edge(4, 1);
    ptr = 5;
    endd();
  }
  ptr = 2;
  int i = 30;
  while (!((k >> i) & 1)) {
    --i;
  }
  if (i == 0) {
    cout << "2\nNY\nYN\n";
    return 0;
  }
  int vert = ptr++;
  add_edge(0, vert);
  add_edge(vert, ptr++);
  add_edge(vert, ptr++);
  map<int, int> lvl;
  lvl[-1] = ptr - 1;
  for (int j = 0; j < i - 1; ++j) {
    int u = ptr++;
    int v = ptr++;
    add_edge(u - 1, v);
    add_edge(u - 2, v);
    add_edge(u - 1, u);
    add_edge(u - 2, u);
    lvl[j] = u;
  }
  add_edge(ptr - 1, 1);
  add_edge(ptr - 2, 1);
  int high = i;
  while (i > 0) {
    --i;
    if (!((k >> i) & 1)) continue;
    add_edge(0, ptr++);
    for (int j = 0; j < high - i - 1; ++j) {
      add_edge(ptr - 1, ptr);
      ptr++;
    }
    add_edge(ptr - 1, lvl[high - i - 2]);
  }
  endd();
  return 0;
}
