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;
vector<pair<int, int>> e;
void add_pair(int a, int b) { e.push_back(make_pair(a, b)); }
int main() {
  bool fl;
  int k;
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY" << endl << "YN";
    return 0;
  }
  vector<int> v;
  int kol;
  while (k > 0) {
    v.push_back(k % 2);
    k /= 2;
  }
  int n = v.size();
  int last = 0;
  fl = false;
  kol = 2;
  for (int i = n; i >= 1; i--) {
    if (fl && v[i - 1] == 0) kol++;
    if (v[i - 1] == 1) {
      if (!fl) {
        fl = true;
        kol += 2 * i - 2;
        i--;
        e.push_back(make_pair(1, 3));
        e.push_back(make_pair(1, 4));
        for (int j = 1; j <= i - 1; j++) {
          add_pair(2 * j + 1, 2 * j + 3);
          add_pair(2 * j + 1, 2 * j + 4);
          add_pair(2 * j + 2, 2 * j + 3);
          add_pair(2 * j + 2, 2 * j + 4);
        }
        add_pair(2 * i + 1, 2);
        add_pair(2 * i + 2, 2);
        last = kol;
        add_pair(last + 1, 2);
        last++;
        i++;
      } else {
        kol++;
        int l = kol;
        if (i == 1) {
          add_pair(1, kol);
          int k1 = kol;
          while (k1 > last) {
            add_pair(k1, k1 - 1);
            k1--;
          }
        } else if (i == n - 1) {
          add_pair(2 * i, last);
          add_pair(2 * i - 1, last);
        } else {
          add_pair(2 * i, kol);
          add_pair(2 * i - 1, kol);
          int k1 = kol;
          while (k1 > last) {
            add_pair(k1, k1 - 1);
            k1--;
          }
          last = kol;
        }
      }
    }
  }
  bool a[1000][1000];
  for (int i = 0; i <= kol + 1; i++)
    for (int j = 0; j <= kol + 1; j++) a[i][j] = false;
  for (int i = 0; i < e.size(); i++) {
    a[e[i].first][e[i].second] = true;
    a[e[i].second][e[i].first] = true;
  }
  cout << kol << endl;
  for (int i = 1; i <= kol; i++)
    for (int j = 1; j <= kol; j++) {
      if (a[i][j])
        cout << "Y";
      else
        cout << "N";
      if (j == kol) cout << endl;
    }
  return 0;
}