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.
#include <bits/stdc++.h>
using namespace std;
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
int g[1200][1200];
int n;
const int PL = 35;
void add_edge(int a, int b) { g[a][b] = g[b][a] = 1; };
void add_ae(pair<int, int> from, pair<int, int> to) {
  add_edge(from.first, to.first);
  add_edge(from.second, to.first);
  add_edge(from.first, to.second);
  add_edge(from.second, to.second);
};
void add_gadget(int size) {
  pair<int, int> pr = make_pair(0, 0);
  for (int i = 0; i < (PL); ++i) {
    pair<int, int> nxt;
    if (i < size) {
      nxt = make_pair(n, n + 1);
      n += 2;
    } else {
      nxt = make_pair(n, n);
      n++;
    };
    add_ae(pr, nxt);
    pr = nxt;
  };
  pair<int, int> nxt = make_pair(1, 1);
  add_ae(pr, nxt);
};
int start_gadget, end_gadget;
void add_gadget() {
  pair<int, int> pr = make_pair(0, 0);
  start_gadget = n;
  for (int i = 0; i < (PL); ++i) {
    pair<int, int> nxt = make_pair(n, n + 1);
    add_ae(pr, nxt);
    n += 2;
    pr = nxt;
  };
  end_gadget = n;
};
int start_line, end_line;
void add_line() {
  start_line = n;
  n++;
  for (int i = 0; i < (PL - 1); ++i) {
    add_edge(n - 1, n);
    n++;
  };
  add_edge(n - 1, 1);
  end_line = n;
};
void add_funny_edges(int i) {
  if (i == 0)
    add_edge(0, start_line);
  else {
    int a = start_gadget + (i - 1) * 2, b = a + 1;
    int c = start_line + i;
    add_ae(make_pair(a, b), make_pair(c, c));
  };
};
int main() {
  n = 2;
  int q;
  scanf("%d", &q);
  add_gadget();
  add_line();
  for (int i = 0; i < (32); ++i) {
    if ((1 << i) & q) {
      add_funny_edges(i);
    };
  };
  printf("%d\n", n);
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      printf(g[i][j] ? "Y" : "N");
    };
    printf("\n");
  };
  return 0;
}
