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;
char g[1003][1003];
int rev[1003];
int dp[1003];
int dfs(int cnode) {
  if (cnode == 1) return 1;
  int &ret = dp[cnode];
  if (ret == -1) {
    ret = 0;
    for (int i = 0; i < 1003; i++) {
      if (g[cnode][i] == 'Y') ret += dfs(i);
    }
  }
  return ret;
}
int main() {
  int i, k;
  cin >> k;
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  vector<pair<int, int> > v;
  string s = "";
  while (k) {
    if ((k & 1)) {
      s += '1';
    } else {
      s += '0';
    }
    k >>= 1;
  }
  int nv = s.size();
  int n = s.size() - 1;
  for (i = 0; i < n; i++) {
    v.push_back(make_pair(i, nv));
    v.push_back(make_pair(nv, i + 1));
    nv++;
    v.push_back(make_pair(i, nv));
    v.push_back(make_pair(nv, i + 1));
    nv++;
  }
  memset(dp, -1, sizeof(dp));
  int j;
  for (i = (int)s.size() - 2; i >= 0; i--) {
    if (s[i] == '1') {
      int pv = i;
      for (j = 1; j < (n - i) * 2; j++) {
        v.push_back(make_pair(pv, nv));
        pv = nv;
        nv++;
      }
      v.push_back(make_pair(pv, n));
    }
  }
  for (i = 0; i < nv; i++) {
    for (j = 0; j < nv; j++) g[i][j] = 'N';
    g[i][j] = 0;
  }
  memset(rev, -1, sizeof(rev));
  rev[0] = 0;
  rev[n] = 1;
  int cnt = 2;
  for (i = 0; i < nv; i++) {
    if (rev[i] == -1) rev[i] = cnt++;
  }
  cout << nv << endl;
  for (i = 0; i < (int)v.size(); i++) {
    g[rev[v[i].first]][rev[v[i].second]] =
        g[rev[v[i].second]][rev[v[i].first]] = 'Y';
  }
  for (i = 0; i < nv; i++) {
    printf("%s\n", g[i]);
  }
  return 0;
}