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 k, sum;
int bin[40], nd = 3;
vector<int> G[1009];
char A[1009][1009];
int main() {
  scanf("%d", &k);
  int sz = 0, mx;
  while (k) {
    bin[sz++] = k % 2;
    k /= 2;
  }
  mx = sz - 1;
  for (int i = sz - 1; i > 0; i--) {
    if (!bin[i]) continue;
    if (i != sz - 1) {
      int a = 1, b = 4;
      for (int j = 0; j < mx - i; j++) {
        G[a].push_back(nd);
        G[nd].push_back(a);
        a = nd;
        nd++;
        b += 2;
      }
      G[a].push_back(b);
      G[b].push_back(a);
      G[a].push_back(b - 1);
      G[b - 1].push_back(a);
      continue;
    }
    int a = 1, b;
    G[a].push_back(nd);
    G[a].push_back(nd + 1);
    G[nd].push_back(a);
    G[nd + 1].push_back(a);
    a = nd, b = nd + 1;
    nd += 2;
    for (int j = 1; j < i; j++) {
      int toa = nd, tob = nd + 1;
      G[a].push_back(toa), G[toa].push_back(a);
      G[a].push_back(tob), G[tob].push_back(a);
      G[b].push_back(toa), G[toa].push_back(b);
      G[b].push_back(tob), G[tob].push_back(b);
      a = toa, b = tob;
      nd += 2;
    }
    G[2].push_back(a);
    G[2].push_back(b);
    G[a].push_back(2);
    G[b].push_back(2);
  }
  if (bin[0]) {
    int a = 1;
    for (int i = 0; i < mx; i++) {
      G[a].push_back(nd);
      G[nd].push_back(a);
      a = nd;
      nd++;
    }
    G[2].push_back(a);
    G[a].push_back(2);
  }
  for (int i = 1; i < nd; i++)
    for (int j = 1; j < nd; j++) A[i][j] = 'N';
  for (int i = 1; i < nd; i++)
    for (int j = 0; j < (int)G[i].size(); j++) A[i][G[i][j]] = 'Y';
  printf("%d\n", nd - 1);
  for (int i = 1; i < nd; i++) {
    for (int j = 1; j < nd; j++) printf("%c", A[i][j]);
    printf("\n");
  }
}
