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;
int mm[1008][1008];
int bit[31];
int main() {
  int k;
  int n, i, d;
  int ni;
  int bi = 0;
  int kk;
  int j;
  scanf("%d", &k);
  if (k == 1) {
    printf("2\n");
    printf("NY\n");
    printf("YN\n");
    return 0;
  }
  kk = k;
  while (kk) {
    bit[bi++] = kk % 2;
    kk = kk / 2;
  }
  int a, b;
  n = 2;
  mm[1][n + 1] = 1;
  mm[n + 1][1] = 1;
  mm[1][n + 2] = 1;
  mm[n + 2][1] = 1;
  mm[1][n + 3] = 1;
  mm[n + 3][1] = 1;
  for (i = 1; i < bi - 1; i++) {
    mm[n + 1][n + 4] = 1;
    mm[n + 4][n + 1] = 1;
    mm[n + 1][n + 5] = 1;
    mm[n + 5][n + 1] = 1;
    mm[n + 2][n + 4] = 1;
    mm[n + 4][n + 2] = 1;
    mm[n + 2][n + 5] = 1;
    mm[n + 5][n + 2] = 1;
    mm[n + 3][n + 6] = 1;
    mm[n + 6][n + 3] = 1;
    n = n + 3;
  }
  mm[n + 1][2] = 1;
  mm[2][n + 1] = 1;
  mm[n + 2][2] = 1;
  mm[2][n + 2] = 1;
  mm[n + 3][2] = 1;
  mm[2][n + 3] = 1;
  n = n + 3;
  for (i = bi - 2; i > 0; i--) {
    if (bit[i] == 1) {
      mm[i * 3][i * 3 + 5] = 1;
      mm[i * 3 + 5][i * 3] = 1;
      mm[i * 3 + 1][i * 3 + 5] = 1;
      mm[i * 3 + 5][i * 3 + 1] = 1;
    }
  }
  if (bit[0] == 0) {
    mm[1][5] = 0;
    mm[5][1] = 0;
  }
  printf("%d\n", n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (mm[i][j] == 1)
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}