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;
bool G[1001][1001];
int b[32];
void addEdge(int i, int j) { G[i][j] = G[j][i] = true; }
int main() {
  int k;
  while (cin >> k) {
    memset(G, 0, sizeof(G));
    memset(b, 0, sizeof(b));
    int temp = k;
    int cnt = 0;
    while (temp) {
      b[cnt++] = temp % 2;
      temp /= 2;
    }
    int mxbit = 0;
    for (int i = 31; i >= 0; i--) {
      if (b[i]) {
        mxbit = i;
        break;
      }
    }
    int layers = mxbit + 2;
    addEdge(1, 3);
    addEdge(1, 4);
    int id = 3;
    int last_id = 3;
    for (int i = 2; i < layers; i++) {
      for (int j = 0; j < i; j++) {
        addEdge(id + j, id + j + i);
        addEdge(id + j, id + i + i);
      }
      id += i;
      last_id = id;
    }
    for (int i = 0; i <= mxbit; i++) {
      if (b[i]) {
        addEdge(last_id + i + 1, 2);
      }
    }
    printf("%d\n", last_id + layers - 1);
    for (int i = 1; i <= last_id + layers - 1; i++) {
      for (int j = 1; j <= last_id + layers - 1; j++) {
        if (G[i][j]) {
          printf("Y");
        } else {
          printf("N");
        }
      }
      printf("\n");
    }
  }
  return 0;
}
