#include <bits/stdc++.h>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<std::vector<int>> suf(n + 1);
  std::vector<int> b, cnt(n + 1), pos(n + 1);
  int xyx = 0;
  int len = 0;
  auto check = [&](int l, int r) {
    for (auto x : suf[r])
      if (x == r - l) return true;
    return false;
  };
  while (n--) {
    int a;
    std::cin >> a;
    ++len;
    suf[len].clear();
    if (len >= 2 && a == b.back()) suf[len].push_back(2);
    for (auto x : suf[len - 1])
      if (x < len - 1 && b[len - x - 2] == a) suf[len].push_back(x + 2);
    b.push_back(a);
    if (!suf[len].empty() &&
        check(len - suf[len][0] * 3 / 2, len - suf[len][0] / 2)) {
      b.resize(len - suf[len][0]);
      len = b.size();
      xyx += 2;
    }
    pos[len] = pos[len - 1];
    cnt[len] = cnt[len - 1];
    if (check(pos[len], len)) {
      pos[len] = (pos[len] + len) / 2;
      ++cnt[len];
    }
    int ans = xyx + cnt[len];
    for (int x = len; x > pos[len];) {
      if (!suf[x].empty() && x - suf[x][0] >= pos[len]) {
        x -= suf[x][0] / 2;
        ++ans;
      } else {
        break;
      }
    }
    std::cout << ans << " \n"[!n];
  }
  return 0;
}

### Prompt

Please formulate 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>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<std::vector<int>> suf(n + 1);
  std::vector<int> b, cnt(n + 1), pos(n + 1);
  int xyx = 0;
  int len = 0;
  auto check = [&](int l, int r) {
    for (auto x : suf[r])
      if (x == r - l) return true;
    return false;
  };
  while (n--) {
    int a;
    std::cin >> a;
    ++len;
    suf[len].clear();
    if (len >= 2 && a == b.back()) suf[len].push_back(2);
    for (auto x : suf[len - 1])
      if (x < len - 1 && b[len - x - 2] == a) suf[len].push_back(x + 2);
    b.push_back(a);
    if (!suf[len].empty() &&
        check(len - suf[len][0] * 3 / 2, len - suf[len][0] / 2)) {
      b.resize(len - suf[len][0]);
      len = b.size();
      xyx += 2;
    }
    pos[len] = pos[len - 1];
    cnt[len] = cnt[len - 1];
    if (check(pos[len], len)) {
      pos[len] = (pos[len] + len) / 2;
      ++cnt[len];
    }
    int ans = xyx + cnt[len];
    for (int x = len; x > pos[len];) {
      if (!suf[x].empty() && x - suf[x][0] >= pos[len]) {
        x -= suf[x][0] / 2;
        ++ans;
      } else {
        break;
      }
    }
    std::cout << ans << " \n"[!n];
  }
  return 0;
}

```