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;
void desperate_optimization(int precision) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
int flow[2005][2005];
long long n;
stack<long long> re;
int main() {
  desperate_optimization(10);
  cin >> n;
  long long lo = 1;
  long long maxi = 0;
  while (lo <= n) {
    lo *= 3LL;
    maxi++;
  }
  maxi--;
  long long tmp = n;
  while (tmp) {
    re.push(tmp % 3);
    tmp /= 3;
  }
  lo = 1;
  long long hi = 2;
  vector<long long> last;
  for (int i = maxi; i >= 0; i--) {
    long long tr = re.top();
    re.pop();
    if (tr == 0) continue;
    long long now = lo;
    for (int j = 0; j < i; j++) {
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      hi++;
      flow[hi - 1][hi] = flow[hi][hi - 1] = 1;
      flow[hi - 2][hi] = flow[hi][hi - 2] = 1;
      flow[hi - 3][hi] = flow[hi][hi - 3] = 1;
      now = hi;
    }
    for (int j = i; j < maxi; j++) {
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      now = hi;
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      now = hi;
    }
    if (tr == 2) {
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      hi++;
      flow[hi - 2][hi] = flow[hi][hi - 2] = 1;
      flow[hi - 1][hi] = flow[hi][hi - 1] = 1;
    } else {
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
      now = hi;
      hi++;
      flow[now][hi] = flow[hi][now] = 1;
    }
    last.push_back(hi);
  }
  for (int i = 0; i < last.size(); i++) {
    int nx = last[i];
    flow[nx][2] = flow[2][nx] = 1;
  }
  cout << hi << '\n';
  for (int i = 1; i <= hi; i++) {
    for (int j = 1; j <= hi; j++) cout << ((flow[i][j] == 1) ? 'Y' : 'N');
    cout << '\n';
  }
  return 0;
}
