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.
#include <bits/stdc++.h>
using namespace std;
int b[1000][1000];
int a[32];
void solve(int x, int y) {
  int last = 0, last1 = 0, last2 = 0;
  for (int i = x - 1; i >= 0; i--) {
    b[y][last] = 1;
    b[last][y] = 1;
    b[y + 1][last] = 1;
    b[last][y + 1] = 1;
    b[y][last1] = 1;
    b[last1][y] = 1;
    b[y + 1][last1] = 1;
    b[last1][y + 1] = 1;
    b[last2][y + 2] = 1;
    b[y + 2][last2] = 1;
    if (a[i] && i != 0) {
      b[y + 2][y + 3] = 1;
      b[y + 3][y + 2] = 1;
      b[y + 2][y + 4] = 1;
      b[y + 4][y + 2] = 1;
    }
    last = y;
    last1 = y + 1;
    last2 = y + 2;
    y += 3;
  }
  if (a[0]) {
    b[1][last2] = 1;
    b[last2][1] = 1;
  }
  b[1][last] = 1;
  b[last][1] = 1;
  b[1][last1] = 1;
  b[last1][1] = 1;
}
void solve2(int y) {
  int last = 0;
  for (int i = 0; i < 32; i++) {
    b[y][last] = 1;
    b[last][y] = 1;
    last = y;
    y++;
  }
  b[1][last] = 1;
  b[last][1] = 1;
  return;
}
int main() {
  int k;
  cin >> k;
  int i = 0;
  while (k > 0) {
    a[i] = k % 2;
    i++;
    k = k / 2;
  }
  memset(b, 0, sizeof(b));
  int j = 2;
  for (int i = 32; i >= 0; i--) {
    if (a[i] == 1) {
      solve(i, j);
      break;
    }
  }
  cout << 100 << endl;
  for (int i = 0; i < 100; i++) {
    for (int I = 0; I < 100; I++) {
      if (b[i][I])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
