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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int bit[1100];
int tot = 0;
void GetBinary(int n) {
  memset(bit, 0, sizeof bit);
  while (n) {
    bit[tot++] = n & 1;
    n >>= 1;
  }
}
int layer[1100][2];
void Construct() {
  layer[1][0] = 1;
  layer[2][0] = 3;
  layer[3][0] = 4;
  layer[3][1] = 5;
  for (int i = 4; i <= 60; i += 2) {
    layer[i][0] = layer[i - 2][0] + 3;
    layer[i + 1][0] = layer[i - 1][0] + 3;
    layer[i + 1][1] = layer[i - 1][1] + 3;
  }
}
int g[1100][1100];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  GetBinary(n);
  Construct();
  memset(g, 0, sizeof g);
  int total = tot << 1;
  layer[total + 1][0] = 2;
  for (int i = 2; i <= total; ++i) {
    if (i == 2) {
      g[layer[i - 1][0]][layer[i][0]] = 1;
      continue;
    }
    if (i % 2 == 1) {
      g[layer[i - 1][0]][layer[i][0]] = 1;
      g[layer[i - 1][0]][layer[i][1]] = 1;
    } else {
      g[layer[i - 1][0]][layer[i][0]] = 1;
      g[layer[i - 1][1]][layer[i][0]] = 1;
    }
  }
  g[2][layer[total][0]] = 1;
  int PathLength = total;
  int ind = tot * 3, indd = tot * 3, end = tot * 3;
  for (int i = 0; i < tot - 1; ++i) {
    if (bit[i]) {
      int pre = layer[(i + 1) * 2][0];
      indd = ind;
      for (int j = 1; j <= PathLength - (i + 1) * 2 - 1; ++j) {
        g[pre][j + indd] = 1;
        pre = j + indd;
        ind++;
      }
      g[pre][end] = 1;
    }
  }
  cout << ind << endl;
  for (int i = 1; i <= ind; ++i) {
    for (int j = 1; j <= ind; ++j) {
      if (i == j)
        cout << "N";
      else if (g[i][j] || g[j][i])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}