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 int inf = 1e9 + 100500;
const int maxn = 1010;
int n;
int a[maxn][maxn];
int b[maxn];
void build_primary_path(int k) {
  int s = 0;
  for (int i = 0; i < (int)(k); ++i) {
    int q = n++;
    int w = n++;
    int e = (i == k - 1 ? 1 : n++);
    a[s][q] = a[s][w] = a[q][e] = a[w][e] = 1;
    b[i] = e;
    s = e;
  }
}
int t;
int d[maxn];
int cnt[maxn];
int q[maxn];
int lq, rq;
void check() {
  for (int i = 0; i < (int)(n); ++i)
    for (int j = 0; j < (int)(n); ++j) a[i][j] |= a[j][i];
  int lq = 0, rq = 1;
  q[0] = 0;
  for (int i = 0; i < (int)(n); ++i) d[i] = 100000;
  d[0] = 0;
  cnt[0] = 1;
  while (lq != rq) {
    int v = q[lq++];
    for (int i = 0; i < (int)(n); ++i)
      if (a[v][i]) {
        if (d[i] > d[v] + 1) {
          d[i] = d[v] + 1;
          q[rq++] = i;
        }
        if (d[i] == d[v] + 1) {
          cnt[i] += cnt[v];
        }
      }
  }
  cout << "cnt[1] = " << cnt[1] << endl;
  assert(cnt[1] == t);
}
int main() {
  cin >> t;
  if (t == 1) {
    cout << "2" << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  n = 2;
  int k = 1;
  while ((1 << (k + 1)) <= t) ++k;
  build_primary_path(k);
  int s = 0;
  for (int i = 0; i < (int)(k); ++i) {
    int q = n++;
    int w = n++;
    a[s][q] = a[q][w] = 1;
    s = w;
    if (t & (1 << (k - i - 1))) {
      a[q][b[i]] = 1;
    }
  }
  printf("%d\n", n);
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(n); ++j) printf("%c", "NY"[a[i][j] | a[j][i]]);
    printf("\n");
  }
  return 0;
}