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;
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (!a) return b;
  return gcd(b % a, a);
}
int ed[2005][2005];
void solve() {
  int k;
  cin >> k;
  int m = k;
  int a[35], i = 0;
  while (m > 0) {
    a[i] = m % 2;
    m /= 2;
    i++;
  }
  int l = i;
  int pre = 2;
  int mini = 1e9 + 1;
  for (int i = 0; i < l; i++)
    if (a[i]) mini = min(mini, i);
  int path[35];
  for (int i = 0; i < l; i++) {
    if (!a[i]) continue;
    int req = 2 * i;
    if (!i) req = 1;
    int st = pre + 1;
    if (i == 0) {
      ed[1][st] = 1;
      for (int j = 1; j < l - 1; j++) {
        path[j] = st + 1;
        ed[st][st + 1] = 1;
        req++;
        st++;
      }
      ed[2][st] = 1;
    } else {
      ed[1][st] = ed[1][st + 1] = 1;
      for (int j = 1; j < i; j++) {
        ed[st][st + 2] = ed[st][st + 3] = ed[st + 1][st + 2] =
            ed[st + 1][st + 3] = 1;
        st = st + 2;
      }
      if (i == l - 1)
        ed[2][st] = ed[2][st + 1] = 1;
      else {
        if (i > mini) {
          ed[st][path[i]] = ed[st + 1][path[i]] = 1;
        } else {
          int x = st + 2;
          req++;
          ed[st][x] = ed[st + 1][x] = 1;
          for (int j = i + 1; j < l - 1; j++) {
            path[j] = x + 1;
            ed[x][x + 1] = 1;
            x++;
            req++;
          }
          ed[2][x] = 1;
        }
      }
    }
    pre += req;
  }
  cout << pre << "\n";
  for (int i = 1; i <= pre; i++) {
    for (int j = 1; j <= pre; j++) {
      if (max(ed[i][j], ed[j][i]))
        cout << "Y";
      else
        cout << "N";
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}