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 n;
int bit[32];
int ans[1005][1005];
int sum = 0;
int id = 3;
void getbit(int x) {
  sum = 0;
  memset(ans, 0, sizeof ans);
  memset(bit, 0, sizeof bit);
  while (x) {
    if (x & 0x01) bit[sum] = 1;
    x >>= 1;
    sum++;
  }
}
void solve() {
  id = 3;
  int mx = -1;
  int mi = INT_MAX;
  int flag = 0;
  for (int i = sum; i >= 0; i--) {
    if (bit[i] && flag == 0) {
      mx = max(i, mx);
      mi = min(i, mi);
      flag = 1;
      ans[1][id] = 1;
      ans[id][1] = 1;
      for (int j = 0; j < i; j++) {
        ans[id][id + 1] = 1;
        ans[id + 1][id] = 1;
        ans[id][id + 2] = 1;
        ans[id + 2][id] = 1;
        ans[id + 1][id + 3] = 1;
        ans[id + 3][id + 1] = 1;
        ans[id + 2][id + 3] = 1;
        ans[id + 3][id + 2] = 1;
        id += 3;
      }
      ans[id][2] = 1;
      ans[2][id] = 1;
    } else if (bit[i] && flag == 1) {
      mi = min(i, mi);
    }
  }
  if (mx != mi) {
    int start = id + 1;
    int tmp = mx - mi;
    ans[3][id + 1] = 1;
    ans[id + 1][3] = 1;
    for (int i = 1; i < 2 * tmp - 1; i++) {
      ans[id + 1][id + 2] = 1;
      ans[id + 2][id + 1] = 1;
      id++;
    }
    ans[id + 1][3 * (tmp + 1)] = 1;
    ans[3 * (tmp + 1)][id + 1] = 1;
    id++;
    for (int i = 0; i < sum; i++) {
      if (i != mx && i != mi && bit[i]) {
        int t = mx - i + 1;
        ans[3 * t][start + 2 * (t - 2)] = 1;
        ans[start + 2 * (t - 2)][3 * t] = 1;
      }
    }
  }
}
void print() {
  printf("%d\n", id);
  for (int i = 1; i <= id; i++) {
    for (int j = 1; j <= id; j++) {
      if (ans[i][j] == 0)
        printf("N");
      else
        printf("Y");
    }
    puts("");
  }
}
int main() {
  while (scanf("%d", &n) == 1) {
    getbit(n);
    solve();
    print();
  }
}