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 MAXN = 1000;
int len;
bool g[MAXN][MAXN];
int a[40];
int main() {
  scanf("%d", &len);
  if (len == 1) {
    puts("2");
    puts("NY");
    puts("YN");
    return 0;
  }
  int bitLen = 0;
  for (int i = 0; i < 30; i++)
    if (((len >> i) & 1) == 1) bitLen = i;
  memset(g, false, sizeof(g));
  int n = 2;
  a[0] = 0;
  for (int i = 0, pre = 0; i < bitLen; i++) {
    g[pre][n] = g[n][pre] = true;
    n++;
    g[pre][n] = g[n][pre] = true;
    n++;
    int nxt = n;
    if (i == bitLen - 1) nxt = 1;
    g[n - 2][nxt] = g[nxt][n - 2] = true;
    g[n - 1][nxt] = g[nxt][n - 1] = true;
    if (i < bitLen - 1) pre = a[i + 1] = n++;
  }
  for (int now = 0; now < bitLen; now++)
    if (((len >> now) & 1) == 1) {
      for (int i = now, pre = a[now]; i < bitLen; i++) {
        g[pre][n] = g[n][pre] = true;
        n++;
        int nxt = n;
        if (i == bitLen - 1) nxt = 1;
        g[n - 1][nxt] = g[nxt][n - 1] = true;
        if (i < bitLen - 1) pre = n++;
      }
    }
  printf("%d\n", n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if (g[i][j])
        putchar('Y');
      else
        putchar('N');
    puts("");
  }
  return 0;
}