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;
char mmp[1234][1234];
bool vis[1234][1234];
long long Po[33];
int num[33];
int main() {
  int k;
  Po[0] = 1;
  for (int i = 1; i <= 32; i++) {
    Po[i] = (long long)2 * Po[i - 1];
  }
  for (int i = 1; i <= 1111; i++) {
    for (int j = 1; j <= 1111; j++) {
      mmp[i][j] = 'N';
    }
  }
  memset(num, 0, sizeof(num));
  scanf("%d", &k);
  int gg = k, cnt = 0;
  while (gg) {
    num[cnt++] = gg % 2;
    gg /= 2;
  }
  cnt--;
  int gm = 3;
  for (int i = 1; i < 2 * cnt; i++) {
    mmp[gm][gm + 1] = mmp[gm + 1][gm] = 'Y';
    gm++;
  }
  int ed = gm;
  mmp[gm][2] = mmp[2][gm] = 'Y';
  mmp[1][gm + 1] = mmp[gm + 1][1] = 'Y';
  gm++;
  int tm = 1;
  for (int i = 0; i < cnt; i++) {
    tm += 2;
    if (num[i]) {
      mmp[gm][tm] = mmp[tm][gm] = 'Y';
    }
    if (i == cnt - 1) {
      mmp[gm][gm + 1] = mmp[gm][gm + 2] = mmp[gm + 1][gm] = mmp[gm + 2][gm] =
          'Y';
      mmp[gm + 1][ed] = mmp[gm + 2][ed] = mmp[ed][gm + 1] = mmp[ed][gm + 2] =
          'Y';
      gm += 2;
      break;
    }
    mmp[gm][gm + 1] = mmp[gm][gm + 2] = mmp[gm + 1][gm] = mmp[gm + 2][gm] = 'Y';
    mmp[gm + 1][gm + 3] = mmp[gm + 2][gm + 3] = mmp[gm + 3][gm + 1] =
        mmp[gm + 3][gm + 2] = 'Y';
    gm += 3;
  }
  if (cnt == 0) {
    gm = 2;
    mmp[1][2] = mmp[2][1] = 'Y';
    mmp[1][1] = mmp[2][2] = 'N';
  }
  printf("%d\n", gm);
  for (int i = 1; i <= gm; i++) {
    for (int j = 1; j <= gm; j++) {
      putchar(mmp[i][j]);
    }
    puts("");
  }
  return 0;
}