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;
int n, k, d, grid[1010][1010], NODES = 0;
vector<int> graph[1010];
int f(int rem) {
  if (!rem) return 0;
  return 1 + f(rem / 2);
}
void h(int u, int cur) {
  if (cur == 1) {
    grid[u][1] = grid[1][u] = 1;
    return;
  }
  int v = ++NODES;
  grid[u][v] = grid[v][u] = 1;
  h(v, cur - 1);
}
void g(int u, int v, int rem, int cur) {
  if (cur == 1) {
    grid[u][1] = grid[1][u] = grid[v][1] = grid[1][v] = 1;
    return;
  }
  int w, x;
  if (rem & 1) {
    w = ++NODES;
    if (NODES == 1) w = ++NODES;
    grid[u][w] = grid[w][u] = grid[v][w] = grid[w][v] = 1;
    h(w, cur - 1);
    rem--;
  }
  w = ++NODES;
  if (NODES == 1) w = ++NODES;
  x = ++NODES;
  grid[u][w] = grid[w][u] = grid[u][x] = grid[x][u] = 1;
  grid[v][w] = grid[w][v] = grid[v][x] = grid[x][v] = 1;
  g(w, x, rem / 2, cur - 1);
}
int main() {
  scanf("%d", &k);
  d = f(k);
  g(0, 0, k, d);
  if (NODES < 2) NODES++;
  printf("%d\n", NODES + 1);
  for (int u = 0; u <= NODES; u++) {
    for (int v = 0; v <= NODES; v++) {
      if (grid[u][v])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
