#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > adj;
char ans[1024][1024];
void construct(int k) {
  if (k == 1) return;
  construct(k >> 1);
  adj.push_back(vector<int>());
  adj.back().push_back(adj.size());
  adj.back().push_back(adj.size() + 1);
  int x = adj.size();
  adj.push_back(vector<int>(1, adj.size() + 2));
  adj.push_back(vector<int>(1, adj.size() + 1));
  if (k & 1) {
    adj[0].push_back(adj.size());
    for (int remain = k >> 2; remain; remain >>= 1) {
      adj.push_back(vector<int>(1, adj.size() + 1));
      adj.push_back(vector<int>(1, adj.size() + 1));
    }
    adj.push_back(vector<int>(1, adj.size() + 1));
    adj[x + 1].back() = adj[x].back() = adj.size();
  }
}
int main() {
  int n, k;
  scanf("%d", &k);
  if (k == 1) {
    puts("2\nNY\nYN");
    return 0;
  }
  construct(k);
  adj.push_back(vector<int>());
  int N = adj.size();
  printf("%d\n", N);
  vector<int> temp;
  temp = adj[1];
  adj[1] = adj.back();
  adj.back() = temp;
  for (int i = 0; i < adj.size(); i++)
    for (int j = 0; j < adj[i].size(); j++)
      if (adj[i][j] == 1)
        adj[i][j] = adj.size() - 1;
      else if (adj[i][j] == adj.size() - 1)
        adj[i][j] = 1;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) ans[i][j] = 'N';
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < adj[i].size(); j++) {
      ans[i][adj[i][j]] = 'Y';
      ans[adj[i][j]][i] = 'Y';
    }
  }
  for (int i = 0; i < N; i++) puts(ans[i]);
  return 0;
}

### Prompt

Generate 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;
vector<vector<int> > adj;
char ans[1024][1024];
void construct(int k) {
  if (k == 1) return;
  construct(k >> 1);
  adj.push_back(vector<int>());
  adj.back().push_back(adj.size());
  adj.back().push_back(adj.size() + 1);
  int x = adj.size();
  adj.push_back(vector<int>(1, adj.size() + 2));
  adj.push_back(vector<int>(1, adj.size() + 1));
  if (k & 1) {
    adj[0].push_back(adj.size());
    for (int remain = k >> 2; remain; remain >>= 1) {
      adj.push_back(vector<int>(1, adj.size() + 1));
      adj.push_back(vector<int>(1, adj.size() + 1));
    }
    adj.push_back(vector<int>(1, adj.size() + 1));
    adj[x + 1].back() = adj[x].back() = adj.size();
  }
}
int main() {
  int n, k;
  scanf("%d", &k);
  if (k == 1) {
    puts("2\nNY\nYN");
    return 0;
  }
  construct(k);
  adj.push_back(vector<int>());
  int N = adj.size();
  printf("%d\n", N);
  vector<int> temp;
  temp = adj[1];
  adj[1] = adj.back();
  adj.back() = temp;
  for (int i = 0; i < adj.size(); i++)
    for (int j = 0; j < adj[i].size(); j++)
      if (adj[i][j] == 1)
        adj[i][j] = adj.size() - 1;
      else if (adj[i][j] == adj.size() - 1)
        adj[i][j] = 1;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) ans[i][j] = 'N';
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < adj[i].size(); j++) {
      ans[i][adj[i][j]] = 'Y';
      ans[adj[i][j]][i] = 'Y';
    }
  }
  for (int i = 0; i < N; i++) puts(ans[i]);
  return 0;
}

```