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 MAXN = 5010;
char g[MAXN][MAXN];
int size = 2;
void add_edge(int u, int v) { g[u][v] = g[v][u] = 'Y'; }
void add_chain(int u, int v, int len) {
  int cur = u;
  for (int i = 0; i < (int)(len); ++i) {
    int next = size++;
    add_edge(cur, next);
    cur = next;
  }
  add_edge(cur, v);
}
int main(int argc, char **argv) {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < (int)(MAXN); ++i) {
    for (int j = 0; j < (int)(MAXN); ++j) {
      g[i][j] = 'N';
    }
  }
  int len = 30;
  vector<int> prev;
  prev.push_back(0);
  for (int j = 0; j < (int)(30); ++j) {
    if (k & 1) {
      for (int j = 0; j < (int)(prev.size()); ++j) {
        add_chain(prev[j], 1, len);
      }
    }
    int v1 = size++;
    int v2 = size++;
    for (int j = 0; j < (int)(prev.size()); ++j) {
      add_edge(prev[j], v1);
      add_edge(prev[j], v2);
    }
    prev.clear();
    prev.push_back(v1);
    prev.push_back(v2);
    --len;
    k >>= 1;
  }
  printf("%d\n", size);
  for (int i = 0; i < (int)(size); ++i) {
    for (int j = 0; j < (int)(size); ++j) {
      printf("%c", g[i][j]);
    }
    printf("\n");
  }
  return 0;
}