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;
long long n;
long long ti;
int cango[2000][2000];
void path(int x, int y) { cango[x][y] = cango[y][x] = 1; }
int main() {
  scanf("%I64d", &n);
  ti = 1;
  while (ti <= n) {
    ti <<= 1;
  }
  int cnt = 2;
  int flag0 = 0;
  int lena = 0;
  int lastbubu = 0;
  int lasti = 1;
  while (ti) {
    if (n & ti) {
      if (flag0 == 0) {
        lena = 0;
        long long tti = ti;
        while (tti) {
          tti >>= 1;
          lena++;
        }
      }
      long long tti = ti;
      int nowlena = 0;
      while (tti) {
        tti >>= 1;
        nowlena++;
      }
      tti = ti;
      int bubu = lena - nowlena;
      int minub = bubu - lastbubu;
      while (minub) {
        minub--;
        cnt++;
        path(cnt, lasti);
        lasti = cnt;
      }
      lastbubu = bubu;
      int lasta, lastb;
      nowlena--;
      if (nowlena == 0) {
        path(lasti, 2);
      } else {
        nowlena--;
        cnt++;
        path(cnt, lasti);
        lasta = cnt;
        cnt++;
        path(cnt, lasti);
        lastb = cnt;
        while (nowlena--) {
          cnt++;
          path(cnt, lasta);
          path(cnt, lastb);
          cnt++;
          path(cnt, lasta);
          path(cnt, lastb);
          lasta = cnt - 1;
          lastb = cnt;
        }
        path(lasta, 2);
        path(lastb, 2);
      }
      flag0 = 1;
    }
    ti >>= 1;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    for (int j = 1; j <= cnt; j++) {
      if (cango[i][j]) {
        putchar('Y');
      } else {
        putchar('N');
      }
    }
    puts("");
  }
  return 0;
}