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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
char s[1005][1005];
int cur = 120, L;
void link(int i, int j) {
  s[i][j] = 'Y';
  s[j][i] = 'Y';
}
void init() {
  for (int(i) = 0; (i) < (int)(1005); (i)++)
    for (int(j) = 0; (j) < (int)(1005); (j)++) s[i][j] = 'N';
  link(1, 3);
  int j = 0;
  for (int x = 3, a = 63; a < 120; a += 2, x++) {
    link(x, a);
    link(a, x + 1);
    link(x, a + 1);
    link(a + 1, x + 1);
    j++;
  }
  assert(j == 29);
  for (int i = 33; i <= 61; i++) link(i, i + 1);
  link(62, 2);
}
void cal(int i) {
  int id = i + 3;
  int pre = id;
  for (int(j) = 0; (j) < (int)(32 - id); (j)++) {
    link(pre, ++cur);
    pre = cur;
  }
  link(pre, id + 30);
}
int main() {
  scanf("%d", &k);
  init();
  for (int i = 29; i >= 0; i--) {
    if (k & (1 << i)) {
      cal(i);
    }
  }
  printf("%d\n", cur);
  for (int i = 1; i <= cur; i++) {
    for (int j = 1; j <= cur; j++) {
      printf("%c", s[i][j]);
    }
    puts("");
  }
  return 0;
}
