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;
int G[1010][1010];
int n, k;
int main() {
  scanf("%d", &k);
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  int mx = 0;
  for (int i = 0; i < 31; ++i)
    if ((k >> i) & 1) mx = i;
  vector<int> cur = {1};
  n = 2;
  int d = 0;
  for (int i = mx - 1; i >= 0; --i) {
    vector<int> tmp;
    n++;
    for (int i = 0; i < cur.size(); ++i) G[cur[i]][n] = G[n][cur[i]] = 1;
    tmp.push_back(n);
    n++;
    for (int i = 0; i < cur.size(); ++i) G[cur[i]][n] = G[n][cur[i]] = 1;
    tmp.push_back(n);
    d++;
    int prev = 1;
    if ((k >> i) & 1) {
      for (int j = 0; j < d; ++j) {
        n++;
        G[prev][n] = G[n][prev] = 1;
        prev = n;
      }
      tmp.push_back(n);
    }
    cur = tmp;
  }
  for (int i = 0; i < cur.size(); ++i) G[cur[i]][2] = G[2][cur[i]] = 1;
  printf("%d\n", n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) printf(G[i][j] ? "Y" : "N");
    printf("\n");
  }
  return 0;
}