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;
struct solution_dag {
  vector<vector<int> > v;
  int path_12;
  solution_dag() {
    v = vector<vector<int> >(4);
    v[0].push_back(2);
    v[2].push_back(3);
    v[3].push_back(1);
    path_12 = 3;
  }
  void pow(int P) {
    int A = v[0][0];
    v[0].clear();
    int B = add_node();
    v[0].push_back(B);
    for (int i = 0; i < P; i++) {
      int C = add_node();
      v[B].push_back(C);
      v[C].push_back(A);
    }
    path_12 += 2;
  }
  void plus_k(int k) {
    int A = v[0][0];
    if (path_12 == 3) {
      for (int i = 0; i < k; i++) {
        int temp = add_node();
        v[A].push_back(temp);
        v[temp].push_back(1);
      }
    } else {
      int prev = add_node();
      for (int i = 0; i < k; i++) {
        int temp = add_node();
        v[A].push_back(temp);
        v[temp].push_back(prev);
      }
      for (int i = 0; i < path_12 - 4; i++) {
        int next = add_node();
        v[prev].push_back(next);
        prev = next;
      }
      v[prev].push_back(1);
    }
  }
  int add_node(void) {
    int res = v.size();
    v.push_back(vector<int>());
    return res;
  }
};
const int MX = 1009;
bool ans[MX][MX];
const int LG = 32;
int k_digs[LG];
const int MOD = 7;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  int k;
  cin >> k;
  for (int i = 0; i < LG; i++) {
    k_digs[i] = k % MOD;
    k /= MOD;
  }
  int it = LG - 1;
  while (k_digs[it] == 0) {
    it--;
  }
  solution_dag D;
  D.plus_k(k_digs[it] - 1);
  it--;
  while (it >= 0) {
    if (k_digs[it] != 0) {
      D.pow(MOD);
      D.plus_k(k_digs[it]);
    } else {
      D.pow(MOD);
    }
    it--;
  }
  int n = D.v.size();
  cout << n << '\n';
  assert(n <= 1000);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) ans[i][j] = false;
  for (int i = 0; i < n; i++) {
    for (size_t j = 0; j < D.v[i].size(); j++) {
      int k = D.v[i][j];
      ans[i][k] = true;
      ans[k][i] = true;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << (ans[i][j] ? 'Y' : 'N');
    cout << '\n';
  }
  return 0;
}