#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int ma[maxn][maxn];
int n;
int bit[50];
int fac(int x) {
  int cnt = 0;
  while (x) {
    bit[++cnt] = x % 2;
    x /= 2;
  }
  return cnt;
}
void unicom(int u, int v) { ma[u][v] = ma[v][u] = 1; }
int main() {
  scanf("%d", &n);
  int len = fac(n);
  int pos = 3;
  unicom(1, 3);
  unicom(1, 4);
  for (int i = 1; i <= 32; ++i) {
    unicom(pos, pos + 2);
    unicom(pos, pos + 3);
    unicom(pos + 1, pos + 2);
    unicom(pos + 1, pos + 3);
    pos += 2;
  }
  int beg = pos + 2;
  pos += 2;
  for (int i = 1; i <= 31; ++i) {
    unicom(pos, pos + 1);
    pos++;
  }
  unicom(2, pos);
  if (bit[1]) unicom(1, beg);
  beg += 2;
  int t = 5;
  for (int i = 2; i <= len; ++i) {
    if (bit[i]) {
      unicom(beg, t);
    }
    beg++;
    t += 2;
  }
  puts("100");
  for (int i = 1; i <= 100; ++i) {
    for (int j = 1; j <= 100; ++j) {
      if (ma[i][j])
        printf("%c", 'Y');
      else
        printf("%c", 'N');
    }
    puts("");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int maxn = 1005;
int ma[maxn][maxn];
int n;
int bit[50];
int fac(int x) {
  int cnt = 0;
  while (x) {
    bit[++cnt] = x % 2;
    x /= 2;
  }
  return cnt;
}
void unicom(int u, int v) { ma[u][v] = ma[v][u] = 1; }
int main() {
  scanf("%d", &n);
  int len = fac(n);
  int pos = 3;
  unicom(1, 3);
  unicom(1, 4);
  for (int i = 1; i <= 32; ++i) {
    unicom(pos, pos + 2);
    unicom(pos, pos + 3);
    unicom(pos + 1, pos + 2);
    unicom(pos + 1, pos + 3);
    pos += 2;
  }
  int beg = pos + 2;
  pos += 2;
  for (int i = 1; i <= 31; ++i) {
    unicom(pos, pos + 1);
    pos++;
  }
  unicom(2, pos);
  if (bit[1]) unicom(1, beg);
  beg += 2;
  int t = 5;
  for (int i = 2; i <= len; ++i) {
    if (bit[i]) {
      unicom(beg, t);
    }
    beg++;
    t += 2;
  }
  puts("100");
  for (int i = 1; i <= 100; ++i) {
    for (int j = 1; j <= 100; ++j) {
      if (ma[i][j])
        printf("%c", 'Y');
      else
        printf("%c", 'N');
    }
    puts("");
  }
  return 0;
}

```