#include <bits/stdc++.h>
using namespace std;
int s, n, k;
char g[1010][1010];
void addEdge(int a, int b) { g[a][b] = g[b][a] = 'Y'; }
int main() {
  scanf("%d", &s);
  if (s == 1) {
    puts("2");
    puts("NY");
    puts("YN");
    return 0;
  }
  k = 30;
  while (((1 << k) & s) == 0) k--;
  memset(g, 'N', sizeof g);
  n = 1 + 2 * (k + 1);
  for (int i = k - 1; i > 0; --i) {
    addEdge(i * 2 + 4, (i - 1) * 2 + 4);
    addEdge(i * 2 + 4, (i - 1) * 2 + 5);
    addEdge(i * 2 + 5, (i - 1) * 2 + 4);
    addEdge(i * 2 + 5, (i - 1) * 2 + 5);
  }
  addEdge(1, 3);
  addEdge((k - 1) * 2 + 4, 3);
  addEdge((k - 1) * 2 + 5, 3);
  addEdge(4, 2);
  addEdge(5, 2);
  for (int i = k - 1; i >= 0; --i) {
    if ((1 << i) & s) {
      int t = k - i, last = 1;
      for (int j = 1; j <= t; ++j) {
        addEdge(last, ++n);
        last = n;
      }
      addEdge(n, i * 2 + 4);
    }
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) putchar(g[i][j]);
    puts("");
  }
}

### 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;
int s, n, k;
char g[1010][1010];
void addEdge(int a, int b) { g[a][b] = g[b][a] = 'Y'; }
int main() {
  scanf("%d", &s);
  if (s == 1) {
    puts("2");
    puts("NY");
    puts("YN");
    return 0;
  }
  k = 30;
  while (((1 << k) & s) == 0) k--;
  memset(g, 'N', sizeof g);
  n = 1 + 2 * (k + 1);
  for (int i = k - 1; i > 0; --i) {
    addEdge(i * 2 + 4, (i - 1) * 2 + 4);
    addEdge(i * 2 + 4, (i - 1) * 2 + 5);
    addEdge(i * 2 + 5, (i - 1) * 2 + 4);
    addEdge(i * 2 + 5, (i - 1) * 2 + 5);
  }
  addEdge(1, 3);
  addEdge((k - 1) * 2 + 4, 3);
  addEdge((k - 1) * 2 + 5, 3);
  addEdge(4, 2);
  addEdge(5, 2);
  for (int i = k - 1; i >= 0; --i) {
    if ((1 << i) & s) {
      int t = k - i, last = 1;
      for (int j = 1; j <= t; ++j) {
        addEdge(last, ++n);
        last = n;
      }
      addEdge(n, i * 2 + 4);
    }
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) putchar(g[i][j]);
    puts("");
  }
}

```