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;
const int MAXN = 1005;
int K, N, Ans, Cnt;
int num[MAXN], dfn[MAXN];
int vis[MAXN][MAXN];
vector<int> P[MAXN];
void Add(int x, int y) { vis[x][y] = vis[y][x] = 1; }
int main() {
  scanf("%d", &K);
  N = 2;
  num[0] = 1;
  for (int i = 30; i >= 0; i--)
    if ((1 << i) & K) {
      Ans = 2 * i;
      K -= (1 << i);
      if (i == 0) {
        Ans = 1;
        Add(1, 2);
        break;
      }
      Add(1, ++N);
      Add(1, ++N);
      for (int j = 1; j < i; j++) {
        Add(N - 1, N + 1);
        Add(N, N + 1);
        N++;
        num[++Cnt] = N;
        Add(N, N + 1);
        Add(N, N + 2);
        N += 2;
      }
      Add(N - 1, 2);
      Add(N, 2);
      num[++Cnt] = 2;
      break;
    }
  if (K) {
    Cnt = 0;
    Add(1, ++N);
    for (int i = 1; i <= Ans - 1; i++) {
      Add(N, N + 1);
      N++;
      if (i % 2 == 1) dfn[++Cnt] = N;
    }
    Add(N, 2);
  }
  for (int i = 30; i >= 0; i--)
    if ((1 << i) & K) Add(num[i], dfn[i + 1]);
  printf("%d\n", N);
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++)
      if (!vis[i][j])
        printf("N");
      else
        printf("Y");
    puts("");
  }
}