#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1, c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
const int N = 1e5 + 4;
int n, top, xyx, ans, a[N], f[N], c[N];
vector<int> vec[N];
inline bool find(int x, int y) {
  for (auto v : vec[x])
    if (v == y) return 1;
  return 0;
}
int main() {
  n = read();
  for (int tt = 1, x; tt <= n; tt++) {
    a[++top] = read();
    vec[top].clear();
    if (top > 1) {
      if (a[top] == a[top - 1]) vec[top].push_back(1);
      for (auto v : vec[top - 1])
        if (a[top - v * 2 - 1] == a[top]) vec[top].push_back(v + 1);
    }
    if (!vec[top].empty() && find(top - vec[top][0], vec[top][0])) {
      top -= vec[top][0] * 2;
      xyx += 2;
    } else {
      f[top] = f[top - 1];
      c[top] = c[top - 1];
      if (!((top - f[top]) & 1) && find(top, (top - f[top]) / 2)) {
        f[top] += (top - f[top]) / 2;
        c[top]++;
      }
    }
    ans = xyx + c[top];
    x = top;
    while (!vec[x].empty() && x - vec[x][0] > f[top]) {
      x -= vec[x][0];
      ans++;
    }
    cout << ans << " ";
  }
  return (0 - 0);
}

### Prompt

Construct a cpp code solution to the problem outlined:

No matter what trouble you're in, don't be afraid, but face it with a smile.

I've made another billion dollars!

— Boboniu

Boboniu has issued his currencies, named Bobo Yuan. Bobo Yuan (BBY) is a series of currencies. Boboniu gives each of them a positive integer identifier, such as BBY-1, BBY-2, etc.

Boboniu has a BBY collection. His collection looks like a sequence. For example:

<image>

We can use sequence a=[1,2,3,3,2,1,4,4,1] of length n=9 to denote it.

Now Boboniu wants to fold his collection. You can imagine that Boboniu stick his collection to a long piece of paper and fold it between currencies:

<image>

Boboniu will only fold the same identifier of currencies together. In other words, if a_i is folded over a_j (1≤ i,j≤ n), then a_i=a_j must hold. Boboniu doesn't care if you follow this rule in the process of folding. But once it is finished, the rule should be obeyed.

A formal definition of fold is described in notes.

According to the picture above, you can fold a two times. In fact, you can fold a=[1,2,3,3,2,1,4,4,1] at most two times. So the maximum number of folds of it is 2.

As an international fan of Boboniu, you're asked to calculate the maximum number of folds. 

You're given a sequence a of length n, for each i (1≤ i≤ n), you need to calculate the maximum number of folds of [a_1,a_2,…,a_i].

Input

The first line contains an integer n (1≤ n≤ 10^5).

The second line contains n integers a_1,a_2,…,a_n (1≤ a_i≤ n).

Output

Print n integers. The i-th of them should be equal to the maximum number of folds of [a_1,a_2,…,a_i].

Examples

Input


9
1 2 3 3 2 1 4 4 1


Output


0 0 0 1 1 1 1 2 2


Input


9
1 2 2 2 2 1 1 2 2


Output


0 0 1 2 3 3 4 4 5


Input


15
1 2 3 4 5 5 4 3 2 2 3 4 4 3 6


Output


0 0 0 0 0 1 1 1 1 2 2 2 3 3 0


Input


50
1 2 4 6 6 4 2 1 3 5 5 3 1 2 4 4 2 1 3 3 1 2 2 1 1 1 2 4 6 6 4 2 1 3 5 5 3 1 2 4 4 2 1 3 3 1 2 2 1 1


Output


0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 5 5 6 7 3 3 3 4 4 4 4 3 3 4 4 4 4 4 5 5 5 5 6 6 6 7 7 8

Note

Formally, for a sequence a of length n, let's define the folding sequence as a sequence b of length n such that:

  * b_i (1≤ i≤ n) is either 1 or -1. 
  * Let p(i)=[b_i=1]+∑_{j=1}^{i-1}b_j. For all 1≤ i<j≤ n, if p(i)=p(j), then a_i should be equal to a_j. 



([A] is the value of boolean expression A. i. e. [A]=1 if A is true, else [A]=0).

Now we define the number of folds of b as f(b)=∑_{i=1}^{n-1}[b_i≠ b_{i+1}].

The maximum number of folds of a is F(a)=max\{ f(b)∣ b  is a folding sequence of a \}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1, c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
const int N = 1e5 + 4;
int n, top, xyx, ans, a[N], f[N], c[N];
vector<int> vec[N];
inline bool find(int x, int y) {
  for (auto v : vec[x])
    if (v == y) return 1;
  return 0;
}
int main() {
  n = read();
  for (int tt = 1, x; tt <= n; tt++) {
    a[++top] = read();
    vec[top].clear();
    if (top > 1) {
      if (a[top] == a[top - 1]) vec[top].push_back(1);
      for (auto v : vec[top - 1])
        if (a[top - v * 2 - 1] == a[top]) vec[top].push_back(v + 1);
    }
    if (!vec[top].empty() && find(top - vec[top][0], vec[top][0])) {
      top -= vec[top][0] * 2;
      xyx += 2;
    } else {
      f[top] = f[top - 1];
      c[top] = c[top - 1];
      if (!((top - f[top]) & 1) && find(top, (top - f[top]) / 2)) {
        f[top] += (top - f[top]) / 2;
        c[top]++;
      }
    }
    ans = xyx + c[top];
    x = top;
    while (!vec[x].empty() && x - vec[x][0] > f[top]) {
      x -= vec[x][0];
      ans++;
    }
    cout << ans << " ";
  }
  return (0 - 0);
}

```