#include <bits/stdc++.h>
using namespace std;
bool edge[1010][1010];
int num;
void add(int s, int e) { edge[s][e] = edge[e][s] = 1; }
void add_path(int s, int e, int x) {
  int a[10];
  for (int i = 0; i < x; i++) a[i] = ++num;
  for (int i = 0; i < x; i++) add(s, a[i]), add(a[i], e);
}
int p[30];
int main() {
  int n, i, j;
  scanf("%d", &n);
  num = 2;
  int tmp = n, cnt = 0;
  while (n) {
    cnt++;
    p[cnt] = n % 4;
    n /= 4;
  }
  for (i = 1; i <= cnt; i++) {
    tmp = ++num;
    add_path(1, tmp, p[i]);
    for (j = 0; j < i - 1; j++) {
      int nxt = ++num;
      add_path(tmp, nxt, 4);
      tmp = nxt;
    }
    for (j = 0; j < 16 - i; j++) {
      int nxt;
      if (j == 15 - i)
        nxt = 2;
      else
        nxt = ++num;
      add_path(tmp, nxt, 1);
      tmp = nxt;
    }
  }
  printf("%d\n", num);
  for (i = 1; i <= num; i++) {
    for (j = 1; j <= num; j++) printf("%c", edge[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
bool edge[1010][1010];
int num;
void add(int s, int e) { edge[s][e] = edge[e][s] = 1; }
void add_path(int s, int e, int x) {
  int a[10];
  for (int i = 0; i < x; i++) a[i] = ++num;
  for (int i = 0; i < x; i++) add(s, a[i]), add(a[i], e);
}
int p[30];
int main() {
  int n, i, j;
  scanf("%d", &n);
  num = 2;
  int tmp = n, cnt = 0;
  while (n) {
    cnt++;
    p[cnt] = n % 4;
    n /= 4;
  }
  for (i = 1; i <= cnt; i++) {
    tmp = ++num;
    add_path(1, tmp, p[i]);
    for (j = 0; j < i - 1; j++) {
      int nxt = ++num;
      add_path(tmp, nxt, 4);
      tmp = nxt;
    }
    for (j = 0; j < 16 - i; j++) {
      int nxt;
      if (j == 15 - i)
        nxt = 2;
      else
        nxt = ++num;
      add_path(tmp, nxt, 1);
      tmp = nxt;
    }
  }
  printf("%d\n", num);
  for (i = 1; i <= num; i++) {
    for (j = 1; j <= num; j++) printf("%c", edge[i][j] ? 'Y' : 'N');
    puts("");
  }
  return 0;
}

```