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;
const int MAX = 1005;
int ma[MAX][MAX];
void setma(int x, int y) { ma[x][y] = ma[y][x] = 1; }
int main() {
  int k;
  scanf("%d", &k);
  int n = 2, big, small, cur = 2, last;
  for (int i = 0; i < 30; i++) {
    if (k & (1 << i)) {
      small = i;
      break;
    }
  }
  for (int i = 30; i >= 0; i--) {
    if (k & (1 << i)) {
      big = i;
      break;
    }
  }
  if (small == 0) small++;
  for (int i = small; i < big; i++) {
    n++;
    if (i + 1 < big)
      setma(cur, cur + 1);
    else
      setma(cur, 1);
    cur++;
    last = cur - 1;
  }
  for (int i = 30; i >= 0; i--) {
    if (k & (1 << i)) {
      int fin = last - (big - (i ? i : 1) - 1);
      if (i == big) fin = 1;
      if (i == 0) {
        n++;
        setma(0, cur);
        setma(cur, fin);
        continue;
      }
      int prv1 = 0, prv2 = 0;
      for (int j = 0; j < i; j++) {
        int a = cur++, b = cur++;
        n += 2;
        setma(prv1, a);
        setma(prv2, b);
        setma(prv1, b);
        setma(prv2, a);
        prv1 = a;
        prv2 = b;
      }
      setma(prv1, fin);
      setma(prv2, fin);
    }
  }
  printf("%d\n", n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf("%c", (ma[i][j] == 1 ? 'Y' : 'N'));
    }
    printf("\n");
  }
  return 0;
}
