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>
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 1e3 + 5;
const double eps = 1e-9;
using namespace std;
int ans[MAXN][MAXN];
int a[MAXN], b[MAXN], c[MAXN][2];
int main() {
  int k;
  scanf("%d", &(k));
  int tmp = 3;
  for (int i = 0; i < 63; i++) a[i] = i + 3;
  ans[65][1] = 1, ans[1][65] = 1;
  for (int i = 0; i < 62; i++) ans[i + 3][i + 4] = 1, ans[i + 4][i + 3] = 1;
  ans[66][2] = 1, ans[2][66] = 1;
  for (int i = 0; i < 31; i++) {
    ans[98 + 2 * i][66 + i] = 1, ans[66 + i][98 + 2 * i] = 1;
    ans[99 + 2 * i][66 + i] = 1, ans[66 + i][99 + 2 * i] = 1;
    ans[98 + 2 * i][67 + i] = 1, ans[67 + i][98 + 2 * i] = 1;
    ans[99 + 2 * i][67 + i] = 1, ans[67 + i][99 + 2 * i] = 1;
  }
  for (int i = 0; i < 32; i++)
    if (k & (1 << i)) ans[2 * i + 3][i + 66] = 1, ans[i + 66][2 * i + 3] = 1;
  printf("%d\n", (160));
  for (int i = 1; i <= 160; i++) {
    for (int j = 1; j <= 160; j++)
      if (ans[i][j])
        printf("Y");
      else
        printf("N");
    printf("\n");
  }
  return 0;
}