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, noww, aa, bb, prea, preb;
bool flag[410][410];
inline void solve() {
  scanf("%d", &K);
  noww = -1;
  prea = preb = K;
  while (1) {
    aa = prea / 2;
    bb = preb - preb / 2;
    if (noww == -1) {
      flag[3][1] = flag[5][1] = 1;
    } else {
      if (aa + aa == prea)
        flag[noww][noww + 4] = flag[noww][noww + 5] = flag[noww + 1][noww + 4] =
            flag[noww + 1][noww + 5] = 1;
      else if (aa + bb == prea) {
        flag[noww][noww + 4] = flag[noww][noww + 6] = flag[noww + 1][noww + 4] =
            flag[noww + 1][noww + 6] = 1;
      } else {
        flag[noww][noww + 6] = flag[noww][noww + 7] = flag[noww + 1][noww + 6] =
            flag[noww + 1][noww + 7] = 1;
      }
      if (aa + aa == preb)
        flag[noww + 2][noww + 4] = flag[noww + 2][noww + 5] =
            flag[noww + 3][noww + 4] = flag[noww + 3][noww + 5] = 1;
      else if (aa + bb == preb) {
        flag[noww + 2][noww + 4] = flag[noww + 2][noww + 6] =
            flag[noww + 3][noww + 4] = flag[noww + 3][noww + 6] = 1;
      } else {
        flag[noww + 2][noww + 6] = flag[noww + 2][noww + 7] =
            flag[noww + 3][noww + 6] = flag[noww + 3][noww + 7] = 1;
      }
    }
    prea = aa;
    preb = bb;
    noww += 4;
    if ((aa == 0 && bb == 1) || (aa == 1 && bb == 1)) break;
  }
  for (int i = noww; i <= noww + 1; i++)
    if (aa == 1) flag[2][i] = 1;
  for (int i = noww + 2; i <= noww + 3; i++)
    if (bb == 1) flag[2][i] = 1;
  printf("%d\n", noww + 3);
  for (int i = 1; i <= noww + 3; i++) {
    for (int j = 1; j <= noww + 3; j++) {
      if (flag[i][j] || flag[j][i])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
int main() {
  solve();
  return 0;
}