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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool G[100][100];
int a[30][2];
void getAll() {
  for (int i = 0; i < 100; i++)
    for (int j = 0; j < 100; j++) G[i][j] = false;
  G[1][3] = true;
  G[1][4] = true;
  G[1][61] = true;
  G[59][2] = true;
  G[60][2] = true;
  G[89][2] = true;
  for (int i = 1; i < 29; i++) {
    int u = i * 2 + 1, v = i * 2 + 2;
    G[u][u + 2] = true;
    G[u][v + 2] = true;
    G[v][u + 2] = true;
    G[v][v + 2] = true;
  }
  for (int i = 61; i <= 88; i++) {
    G[i][i + 1] = true;
    int a = i - 59;
    G[i][a * 2 + 1] = true;
    G[i][a * 2 + 2] = true;
  }
}
void putOff(int i) {
  if (i == 0) {
    G[89][2] = false;
    return;
  } else if (i == 29) {
    G[1][3] = false;
    G[1][4] = false;
  } else {
    int v = 89 - i;
    int a = v - 59;
    G[v][a * 2 + 1] = false;
    G[v][a * 2 + 2] = false;
  }
}
int main() {
  int K;
  scanf("%d", &K);
  getAll();
  for (int i = 0; i <= 29; i++) {
    if (((K >> i) & 1) == 0) {
      putOff(i);
    }
  }
  printf("89\n");
  for (int i = 1; i <= 89; i++) {
    for (int j = 1; j <= 89; j++) {
      if (G[i][j] || G[j][i])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}