#include <bits/stdc++.h>
using namespace std;
using ai = int[102333];
using vi = vector<int>;
ai a, q, c, p;
vi v[102333];
int main() {
  int n, S = 0, i = 0;
  scanf("%d", &n);
  q[0] = 1;
  for (int _ = 1; _ <= n; ++_) {
    ++i;
    scanf("%d", a + i);
    v[i].clear();
    if (i > 1) {
      if (a[i] == a[i - 1]) v[i].push_back(2);
      for (int x : v[i - 1])
        if (a[i] == a[i - x - 1]) v[i].push_back(x + 2);
    }
    auto is_pal = [](int L, int R) {
      for (int x : v[R])
        if (x == R - L + 1) return true;
      return false;
    };
    int x = v[i].size() ? v[i][0] : 0;
    if (v[i].size() && is_pal(i - x / 2 - x + 1, i - x / 2))
      i -= x, S += 2;
    else {
      p[i] = x;
      q[i] = q[i - 1], c[i] = c[i - 1];
      if (is_pal(q[i], i)) q[i] += (i - q[i] + 1) / 2, ++c[i];
    }
    int res = S + c[i];
    int cur = i;
    while (p[cur] && q[i] <= cur - p[cur] / 2) cur -= p[cur] / 2, ++res;
    printf("%d%c", res, " \n"[_ == n]);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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;
using ai = int[102333];
using vi = vector<int>;
ai a, q, c, p;
vi v[102333];
int main() {
  int n, S = 0, i = 0;
  scanf("%d", &n);
  q[0] = 1;
  for (int _ = 1; _ <= n; ++_) {
    ++i;
    scanf("%d", a + i);
    v[i].clear();
    if (i > 1) {
      if (a[i] == a[i - 1]) v[i].push_back(2);
      for (int x : v[i - 1])
        if (a[i] == a[i - x - 1]) v[i].push_back(x + 2);
    }
    auto is_pal = [](int L, int R) {
      for (int x : v[R])
        if (x == R - L + 1) return true;
      return false;
    };
    int x = v[i].size() ? v[i][0] : 0;
    if (v[i].size() && is_pal(i - x / 2 - x + 1, i - x / 2))
      i -= x, S += 2;
    else {
      p[i] = x;
      q[i] = q[i - 1], c[i] = c[i - 1];
      if (is_pal(q[i], i)) q[i] += (i - q[i] + 1) / 2, ++c[i];
    }
    int res = S + c[i];
    int cur = i;
    while (p[cur] && q[i] <= cur - p[cur] / 2) cur -= p[cur] / 2, ++res;
    printf("%d%c", res, " \n"[_ == n]);
  }
  return 0;
}

```