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<int> E[2005];
bool M[2005][2005];
void add_edge(int u, int v) {
  E[u].push_back(v);
  E[v].push_back(u);
}
int main() {
  int n, k, lvl;
  while (cin >> k) {
    memset(M, 0, sizeof(M));
    n = 1;
    lvl = 0;
    for (int i = 0; i < 2005; i++) E[i].clear();
    vector<int> v;
    while (k) {
      v.push_back(k % 11);
      k /= 11;
    }
    int nv = v.size();
    reverse(v.begin(), v.end());
    for (int j = 1; j <= v[0]; ++j) {
      add_edge(n, n + j), add_edge(n + j, n + v[0] + 1);
    }
    n += v[0] + 1;
    lvl += 2;
    for (int i = 1; i < nv; ++i) {
      k = v[i];
      for (int j = 1; j <= 11; ++j) {
        add_edge(n, n + j), add_edge(n + j, n + 11 + 1);
      }
      n += 11 + 1;
      lvl += 2;
      if (k) {
        for (int j = 0; j < k; ++j) {
          add_edge(1, n + j * lvl + 1);
        }
        for (int j = 0; j < k; ++j)
          for (int L = 1; L <= lvl - 1; ++L) {
            add_edge(n + j * lvl + L, n + j * lvl + L + 1);
          }
        for (int j = 1; j <= k; ++j) add_edge(n + k * lvl + 1, n + j * lvl);
        add_edge(n, n + k * lvl + 1);
        n += k * lvl + 1;
        lvl++;
      }
    }
    add_edge(n, 0);
    n++;
    for (int i = 0; i < n; i++)
      for (typeof E[i].begin() v = E[i].begin(); v != E[i].end(); v++)
        M[i][*v] = M[*v][i] = 1;
    printf("%d\n", n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        printf("%c%s", M[i][j] ? 'Y' : 'N', j + 1 == n ? "\n" : "");
  }
}