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 k, cnt, point, g;
int q[10000], v[5100][5100], a[1000];
int main() {
  scanf("%d", &k);
  for (; k; k /= 2) a[++g] = k & 1;
  point = 2;
  v[1][2 + g] = v[2 + g][1] = 1;
  int now1 = 2 + g, now2 = 2 + g;
  for (int i = 1; i < g; ++i) {
    ++point;
    v[point][now1] = v[now1][point] = v[now1][point + g] = v[point + g][now1] =
        1;
    v[point + g][now2] = v[now2][point + g] = v[point][now2] = v[now2][point] =
        1;
    now1 = point, now2 = point + g;
  }
  v[now1][2] = v[2][now1] = v[now2][2] = v[2][now2] = 1;
  point += g;
  int now = 1;
  for (int i = g - 1; i >= 1; --i) {
    ++point;
    v[point][now] = v[now][point] = 1;
    now = point;
    if (a[i]) v[2 + g - i][point] = v[point][2 + g - i] = 1;
  }
  printf("%d\n", point);
  for (int i = 1; i <= point; ++i, printf("\n"))
    for (int j = 1; j <= point; ++j) printf("%c", v[i][j] ? 'Y' : 'N');
}