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;
void set(int& n, int first) { n = n | (1 << first); }
void reset(int& n, int first) { n = n & ~(1 << first); }
bool check(int n, int first) { return (bool)(n & (1 << first)); }
const int mx = 110;
char s[mx][mx];
int main() {
  int i, j, k, n;
  while (scanf("%d", &k) == 1) {
    n = 0;
    while ((1LL << (n + 1)) <= k) n++;
    n++;
    for (i = 0; i < mx; i++)
      for (j = 0; j < mx; j++) s[i][j] = 'N';
    for (i = 3 * n + 2, j = i - n; i > 2 * n + 3; i--, j--) {
      s[i][i - 1] = s[i][j - 1] = s[j][i - 1] = s[j][j - 1] = 'Y';
      s[i - 1][i] = s[j - 1][i] = s[i - 1][j] = s[j - 1][j] = 'Y';
    }
    for (i = n + 2; i > 3; i--) s[i][i - 1] = s[i - 1][i] = 'Y';
    s[2][2 * n + 3] = s[2 * n + 3][2] = 'Y';
    s[2][n + 3] = s[n + 3][2] = 'Y';
    s[1][n + 2] = s[n + 2][1] = 'Y';
    for (i = n + 2; i >= 3; i--) {
      if ((1 << (i - 3)) <= k) {
        k -= (1 << (i - 3));
        s[i][i + n] = s[i + n][i] = 'Y';
      }
    }
    n = 3 * n + 2;
    printf("%d\n", n);
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) printf("%c", s[i][j]);
      printf("\n");
    }
  }
}