Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double pi = acos(-1.0);
const double eps = (double)1e-9;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 1005;
char c[N][N];
int k, mx, sz, v[30][3];
inline void add_edge(int x, int y) { c[x][y] = c[y][x] = 'Y'; }
int main() {
  for (int i = 1; i < N; i++)
    for (int j = 1; j < N; j++) c[i][j] = 'N';
  scanf("%d", &k);
  sz = 2;
  for (int i = 29; i >= 1; i--) {
    if ((1 << i) <= k) {
      mx = i + 1;
      int last1 = ++sz;
      int last2 = ++sz;
      add_edge(1, last1);
      add_edge(1, last2);
      v[1][1] = last1;
      v[1][2] = last2;
      for (int j = 1; j < i; j++) {
        int cur1 = ++sz;
        int cur2 = ++sz;
        add_edge(last1, cur1);
        add_edge(last1, cur2);
        add_edge(last2, cur1);
        add_edge(last2, cur2);
        last1 = cur1;
        last2 = cur2;
        v[j + 1][1] = last1;
        v[j + 1][2] = last2;
      }
      add_edge(last1, 2);
      add_edge(last2, 2);
      k ^= (1 << i);
      break;
    }
  }
  for (int i = 29; i >= 1; i--) {
    if ((1 << i) <= k) {
      int last = ++sz;
      add_edge(v[i][1], last);
      add_edge(v[i][2], last);
      for (int j = 1; j <= mx - i - 2; j++) {
        int cur = ++sz;
        add_edge(last, cur);
        last = cur;
      }
      add_edge(last, 2);
      k ^= (1 << i);
    }
  }
  if (k) {
    int last = 1;
    for (int i = 1; i <= mx - 1; i++) {
      int cur = ++sz;
      add_edge(last, cur);
      last = cur;
    }
    add_edge(last, 2);
    k--;
  }
  printf("%d\n", sz);
  for (int i = 1; i <= sz; i++) {
    for (int j = 1; j <= sz; j++) printf("%c", c[i][j]);
    puts("");
  }
  return 0;
}