#include <bits/stdc++.h>
using namespace std;
int n;
char s[1005][1005];
void init() {
  for (int i = 0; i < 1000; i++)
    for (int j = 0; j < 1000; j++) s[i][j] = 'N';
  scanf("%d", &n);
}
void addEdge(int x, int y) { s[x][y] = s[y][x] = 'Y'; }
vector<int> v;
void process() {
  if (n == 1) {
    printf("2\nNY\nYN\n");
    return;
  }
  while (n != 1) {
    if (n % 2) v.push_back(1);
    v.push_back(2);
    n /= 2;
  }
  reverse(v.begin(), v.end());
  int c = 1, last = 0, len = 0;
  for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
    if (*it == 1) {
      addEdge(0, c);
      for (int i = 1; i < len - 1; i++) addEdge(c, c + 1), c++;
      addEdge(c, last), c++;
    } else {
      addEdge(last, c);
      addEdge(last, c + 1);
      addEdge(c, c + 2);
      addEdge(c + 1, c + 2);
      last = c + 2;
      c += 3;
      len += 2;
    }
  }
  for (int i = 0; i < c; i++)
    if (i != 1 && i != last) {
      swap(s[1][i], s[last][i]);
      swap(s[i][1], s[i][last]);
    }
  printf("%d\n", c);
  for (int i = 0; i < c; i++) {
    s[i][c] = '\0';
    puts(s[i]);
  }
}
int main() {
  init();
  process();
  return 0;
}

### Prompt

Please formulate 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 n;
char s[1005][1005];
void init() {
  for (int i = 0; i < 1000; i++)
    for (int j = 0; j < 1000; j++) s[i][j] = 'N';
  scanf("%d", &n);
}
void addEdge(int x, int y) { s[x][y] = s[y][x] = 'Y'; }
vector<int> v;
void process() {
  if (n == 1) {
    printf("2\nNY\nYN\n");
    return;
  }
  while (n != 1) {
    if (n % 2) v.push_back(1);
    v.push_back(2);
    n /= 2;
  }
  reverse(v.begin(), v.end());
  int c = 1, last = 0, len = 0;
  for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
    if (*it == 1) {
      addEdge(0, c);
      for (int i = 1; i < len - 1; i++) addEdge(c, c + 1), c++;
      addEdge(c, last), c++;
    } else {
      addEdge(last, c);
      addEdge(last, c + 1);
      addEdge(c, c + 2);
      addEdge(c + 1, c + 2);
      last = c + 2;
      c += 3;
      len += 2;
    }
  }
  for (int i = 0; i < c; i++)
    if (i != 1 && i != last) {
      swap(s[1][i], s[last][i]);
      swap(s[i][1], s[i][last]);
    }
  printf("%d\n", c);
  for (int i = 0; i < c; i++) {
    s[i][c] = '\0';
    puts(s[i]);
  }
}
int main() {
  init();
  process();
  return 0;
}

```