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 AdjMat[1005][1005];
int d[1005], power[35];
int idx = 3;
inline int NewNode() { return idx++; }
int main() {
  int k, a, b, c = 1;
  cin >> k;
  int prv = 1, MSB = 0;
  d[1] = 0;
  for (int i = 30; i >= 0 && k > 1; i--) {
    if (((1 << i) & k) == 0) continue;
    prv = NewNode();
    AdjMat[1][prv] = 1;
    d[prv] = 1;
    for (int j = 0; j < i; j++) {
      a = NewNode(), b = NewNode(), c = NewNode();
      AdjMat[prv][a] = 1;
      AdjMat[prv][b] = 1;
      AdjMat[a][c] = 1;
      AdjMat[b][c] = 1;
      power[i - j - 1] = c;
      d[a] = 1 + d[prv];
      d[b] = d[a];
      d[c] = 1 + d[a];
      prv = c;
    }
    AdjMat[c][2] = 1;
    MSB = i;
    break;
  }
  for (int i = MSB - 1; i >= 0; i--) {
    if (((1 << i) & k) == 0) continue;
    prv = 1;
    for (int j = 1; j < d[power[i]]; j++) {
      int t = NewNode();
      AdjMat[prv][t] = 1;
      prv = t;
    }
    AdjMat[prv][power[i]] = 1;
  }
  if (k == 1) {
    idx = 3;
    AdjMat[1][2] = 1;
  }
  printf("%d\n", idx - 1);
  for (int i = 1; i < idx; i++) {
    for (int j = 1; j < idx; j++) {
      AdjMat[i][j] |= AdjMat[j][i];
      AdjMat[j][i] |= AdjMat[i][j];
      putchar(AdjMat[i][j] ? 'Y' : 'N');
    }
    puts("");
  }
  return 0;
}
