#include <bits/stdc++.h>
using namespace std;
int n, k;
bool vis[1005][1005];
char s[100];
int main() {
  scanf("%d", &k);
  sprintf(s, "%d", k);
  int len = strlen(s);
  memset(vis, 0, sizeof(vis));
  int now = 2, last;
  for (int i = 0; i < len; i++) {
    int tmp = s[i] - '0';
    if (!tmp) continue;
    for (int j = 1; j <= tmp; j++) {
      vis[1][now + j] = vis[now + j][1] = 1;
    }
    last = now;
    now += tmp;
    int need = len - i - 1;
    for (int j = 1; j <= need; j++) {
      last = now;
      if (j == 1) {
        for (int p = now - tmp + 1; p <= now; p++)
          for (int q = now + 1; q <= now + 10; q++) vis[p][q] = vis[q][p] = 1;
      } else {
        for (int p = now - 9; p <= now; p++)
          for (int q = now + 1; q <= now + 10; q++) vis[p][q] = vis[q][p] = 1;
      }
      now += 10;
    }
    if (i == 0) {
      for (int q = last + 1; q <= now; q++) vis[q][2] = vis[2][q] = 1;
    }
    for (int j = 1; j <= i; j++) {
      if (j == 1) {
        for (int p = last + 1; p <= now; p++)
          vis[p][now + 1] = vis[now + 1][p] = 1;
        now++;
      } else {
        vis[now][now + 1] = vis[now + 1][now] = 1;
        now++;
      }
      if (j == i) vis[now][2] = vis[2][now] = 1;
    }
  }
  printf("%d\n", now);
  for (int i = 1; i <= now; i++) {
    for (int j = 1; j <= now; j++) {
      if (vis[i][j])
        printf("Y");
      else
        printf("N");
    }
    puts("");
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 n, k;
bool vis[1005][1005];
char s[100];
int main() {
  scanf("%d", &k);
  sprintf(s, "%d", k);
  int len = strlen(s);
  memset(vis, 0, sizeof(vis));
  int now = 2, last;
  for (int i = 0; i < len; i++) {
    int tmp = s[i] - '0';
    if (!tmp) continue;
    for (int j = 1; j <= tmp; j++) {
      vis[1][now + j] = vis[now + j][1] = 1;
    }
    last = now;
    now += tmp;
    int need = len - i - 1;
    for (int j = 1; j <= need; j++) {
      last = now;
      if (j == 1) {
        for (int p = now - tmp + 1; p <= now; p++)
          for (int q = now + 1; q <= now + 10; q++) vis[p][q] = vis[q][p] = 1;
      } else {
        for (int p = now - 9; p <= now; p++)
          for (int q = now + 1; q <= now + 10; q++) vis[p][q] = vis[q][p] = 1;
      }
      now += 10;
    }
    if (i == 0) {
      for (int q = last + 1; q <= now; q++) vis[q][2] = vis[2][q] = 1;
    }
    for (int j = 1; j <= i; j++) {
      if (j == 1) {
        for (int p = last + 1; p <= now; p++)
          vis[p][now + 1] = vis[now + 1][p] = 1;
        now++;
      } else {
        vis[now][now + 1] = vis[now + 1][now] = 1;
        now++;
      }
      if (j == i) vis[now][2] = vis[2][now] = 1;
    }
  }
  printf("%d\n", now);
  for (int i = 1; i <= now; i++) {
    for (int j = 1; j <= now; j++) {
      if (vis[i][j])
        printf("Y");
      else
        printf("N");
    }
    puts("");
  }
  return 0;
}

```