#include <bits/stdc++.h>
using namespace std;
int tot;
int vis[1005][1005];
int a[50], b[50];
void addedge(int a, int b) { vis[a][b] = vis[b][a] = 1; }
void init() {
  for (int i = 0; i <= 29; i++) a[i] = 2 + i * 3;
  for (int i = 29; i >= 1; i--) {
    addedge(a[i], a[i] - 1);
    addedge(a[i], a[i] - 2);
    addedge(a[i - 1], a[i] - 1);
    addedge(a[i - 1], a[i] - 2);
  }
}
void link(int a, int b, int num) {
  if (num == 0) {
    addedge(a, b);
    return;
  }
  addedge(a, ++tot);
  while (--num) addedge(tot, tot + 1), tot++;
  addedge(tot, b);
}
int main() {
  int k;
  scanf("%d", &k);
  init();
  int cnt = 0;
  for (int i = 31; i >= 0; i--) {
    if (k & (1 << i)) {
      b[cnt++] = i;
    }
  }
  tot = 89;
  for (int i = 0; i < cnt; i++) {
    int t = 2 * (29 - b[i]);
    link(1, a[b[i]], t);
  }
  printf("%d\n", tot);
  for (int i = 1; i <= tot; i++) {
    for (int j = 1; j <= tot; j++) {
      if (vis[i][j] == 1)
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int tot;
int vis[1005][1005];
int a[50], b[50];
void addedge(int a, int b) { vis[a][b] = vis[b][a] = 1; }
void init() {
  for (int i = 0; i <= 29; i++) a[i] = 2 + i * 3;
  for (int i = 29; i >= 1; i--) {
    addedge(a[i], a[i] - 1);
    addedge(a[i], a[i] - 2);
    addedge(a[i - 1], a[i] - 1);
    addedge(a[i - 1], a[i] - 2);
  }
}
void link(int a, int b, int num) {
  if (num == 0) {
    addedge(a, b);
    return;
  }
  addedge(a, ++tot);
  while (--num) addedge(tot, tot + 1), tot++;
  addedge(tot, b);
}
int main() {
  int k;
  scanf("%d", &k);
  init();
  int cnt = 0;
  for (int i = 31; i >= 0; i--) {
    if (k & (1 << i)) {
      b[cnt++] = i;
    }
  }
  tot = 89;
  for (int i = 0; i < cnt; i++) {
    int t = 2 * (29 - b[i]);
    link(1, a[b[i]], t);
  }
  printf("%d\n", tot);
  for (int i = 1; i <= tot; i++) {
    for (int j = 1; j <= tot; j++) {
      if (vis[i][j] == 1)
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}

```