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;
const int N = 1000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
int cnt = 0, len, s, f;
vector<int> g[N];
void solve(int k) {
  if (k == 1) {
    g[0].push_back(1);
    g[1].push_back(0);
    s = 0;
    f = 1;
    len = 1;
    cnt += 2;
  } else if (k % 2 == 0) {
    solve(k / 2);
    g[cnt].push_back(s);
    g[s].push_back(cnt);
    g[cnt + 1].push_back(s);
    g[s].push_back(cnt + 1);
    g[cnt + 2].push_back(cnt);
    g[cnt].push_back(cnt + 2);
    g[cnt + 2].push_back(cnt + 1);
    g[cnt + 1].push_back(cnt + 2);
    s = cnt + 2;
    len += 2;
    cnt += 3;
  } else {
    solve(k - 1);
    for (int i = 0; i < len; i++) {
      if (i == len - 1) {
        g[cnt - 1].push_back(f);
        g[f].push_back(cnt - 1);
      } else {
        g[cnt - 1].push_back(cnt);
        g[cnt].push_back(cnt - 1);
        cnt++;
      }
    }
  }
}
char ans[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  int k;
  cin >> k;
  solve(k);
  int id[cnt];
  int curr = 2;
  for (int i = 0; i < cnt; i++)
    if (i == s)
      id[i] = 0;
    else if (i == f)
      id[i] = 1;
    else
      id[i] = curr++;
  for (int i = 0; i < cnt; i++)
    for (int j = 0; j < cnt; j++) ans[i][j] = 'N';
  for (int i = 0; i < cnt; i++)
    for (int j : g[i]) ans[id[i]][id[j]] = 'Y';
  cout << cnt << "\n";
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < cnt; j++) cout << ans[i][j];
    cout << "\n";
  }
  return 0;
}