#include <bits/stdc++.h>
using namespace std;
const int N = 1111;
const int64_t LIM = 1000000000;
int a[N][N];
int last = 2;
int len = 20;
void connect(int x, int y) { a[y][x] = a[x][y] = 1; }
void connect(int x, int y, int len, int p) {
  if (a[x][y]) {
    connect(0, 1, p, p);
    return;
  }
  if (len == 0) {
    connect(x, y);
    return;
  }
  connect(x, last);
  for (int i = 0; i < len - 1; i++) {
    connect(last, last + 1);
    last++;
  }
  connect(last, y);
  last++;
}
void build(int p, int k) {
  if (p == 0) {
    connect(0, 1);
    return;
  }
  connect(0, last);
  connect(0, last + 1);
  for (int i = 0; i < p - 1; i++) {
    connect(last, last + 2);
    connect(last, last + 3);
    connect(last + 1, last + 2);
    connect(last + 1, last + 3);
    last += 2;
  }
  connect(last, last + 2);
  connect(last + 1, last + 2);
  last += 2;
  connect(last, 1);
  last++;
  for (int i = 0; i < p; i++) {
    if (k & (1 << i)) {
      connect((i + 1) * 2, 1, p - i, p);
    }
  }
}
int main() {
  int k;
  scanf("%d", &k);
  int mx = 0;
  for (int i = 31; i > 0; i--) {
    if (k & (1 << i)) {
      mx = i;
      break;
    }
  }
  build(mx, k);
  printf("%d\n", last);
  for (int i = 0; i < last; i++) {
    for (int j = 0; j < last; j++) {
      printf(a[i][j] ? "Y" : "N");
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
const int N = 1111;
const int64_t LIM = 1000000000;
int a[N][N];
int last = 2;
int len = 20;
void connect(int x, int y) { a[y][x] = a[x][y] = 1; }
void connect(int x, int y, int len, int p) {
  if (a[x][y]) {
    connect(0, 1, p, p);
    return;
  }
  if (len == 0) {
    connect(x, y);
    return;
  }
  connect(x, last);
  for (int i = 0; i < len - 1; i++) {
    connect(last, last + 1);
    last++;
  }
  connect(last, y);
  last++;
}
void build(int p, int k) {
  if (p == 0) {
    connect(0, 1);
    return;
  }
  connect(0, last);
  connect(0, last + 1);
  for (int i = 0; i < p - 1; i++) {
    connect(last, last + 2);
    connect(last, last + 3);
    connect(last + 1, last + 2);
    connect(last + 1, last + 3);
    last += 2;
  }
  connect(last, last + 2);
  connect(last + 1, last + 2);
  last += 2;
  connect(last, 1);
  last++;
  for (int i = 0; i < p; i++) {
    if (k & (1 << i)) {
      connect((i + 1) * 2, 1, p - i, p);
    }
  }
}
int main() {
  int k;
  scanf("%d", &k);
  int mx = 0;
  for (int i = 31; i > 0; i--) {
    if (k & (1 << i)) {
      mx = i;
      break;
    }
  }
  build(mx, k);
  printf("%d\n", last);
  for (int i = 0; i < last; i++) {
    for (int j = 0; j < last; j++) {
      printf(a[i][j] ? "Y" : "N");
    }
    printf("\n");
  }
  return 0;
}

```