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 ans[1005][1005];
int startID1[35];
int startID2[35];
int helpID[35];
int id = 3;
void createWays() {
  startID1[0] = 2;
  startID2[0] = 2;
  id = 3;
  for (int i = 1; i <= 30; i++) {
    startID1[i] = id;
    startID2[i] = id + 1;
    ans[startID1[i]][startID1[i - 1]] = 1;
    ans[startID2[i]][startID2[i - 1]] = 1;
    ans[startID1[i]][startID2[i - 1]] = 1;
    ans[startID2[i]][startID1[i - 1]] = 1;
    id = id + 2;
  }
  for (int i = 30; i >= 0; i--) {
    if (i == 30) {
      ans[1][id] = 1;
      helpID[i] = id;
      id++;
    } else {
      ans[id - 1][id] = 1;
      helpID[i] = id;
      id++;
    }
  }
  for (int i = 1; i <= id; i++)
    for (int j = 1; j <= id; j++) ans[i][j] = ans[i][j] | ans[j][i];
}
int main() {
  createWays();
  int k;
  cin >> k;
  int i = 0;
  while (i < 31) {
    if ((1 << i) & k) {
      ans[startID1[i]][helpID[i]] = 1;
      ans[helpID[i]][startID1[i]] = 1;
      ans[startID2[i]][helpID[i]] = 1;
      ans[helpID[i]][startID2[i]] = 1;
    }
    i++;
  }
  cout << id << "\n";
  for (int i = 1; i <= id; i++) {
    for (int j = 1; j <= id; j++) {
      cout << (ans[i][j] ? 'Y' : 'N');
    }
    cout << "\n";
  }
  return 0;
}
