#include <bits/stdc++.h>
int main() {
  using namespace std;
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int N;
  cin >> N;
  vector<int> A(N);
  for (auto& a : A) {
    cin >> a;
    a--;
  }
  vector<int> ans;
  ans.reserve(N);
  vector<int> curString;
  curString.reserve(N);
  vector<int> numSuffFolds;
  numSuffFolds.reserve(N + 1);
  vector<vector<int>> suffixPalindromes;
  suffixPalindromes.reserve(N + 1);
  auto reset = [&]() {
    curString.clear();
    numSuffFolds.clear();
    numSuffFolds.push_back(0);
    suffixPalindromes.clear();
    suffixPalindromes.emplace_back();
  };
  auto add_char = [&](int v) -> void {
    int r = int(curString.size()) + 1;
    vector<int> nSuffixPalindromes;
    nSuffixPalindromes.reserve(suffixPalindromes.back().size() + 1);
    if (!curString.empty() && curString.back() == v)
      nSuffixPalindromes.push_back(r - 1);
    for (int m : suffixPalindromes.back()) {
      if (2 * m - r >= 0 && v == curString[2 * m - r]) {
        nSuffixPalindromes.push_back(m);
      }
    }
    curString.push_back(v);
    suffixPalindromes.push_back(nSuffixPalindromes);
    numSuffFolds.push_back(nSuffixPalindromes.empty()
                               ? 0
                               : numSuffFolds[nSuffixPalindromes[0]] + 1);
  };
  int curFolds = 0;
  reset();
  for (int v : A) {
    add_char(v);
    int r = int(curString.size());
    if (suffixPalindromes.back().size() == 1) {
      int m = suffixPalindromes.back()[0];
      int l = m - (r - m);
      if (!suffixPalindromes[m].empty() && suffixPalindromes[m][0] == l) {
        curFolds += 2;
        curString.resize(l);
        numSuffFolds.resize(l + 1);
        suffixPalindromes.resize(l + 1);
      } else if (l == 0) {
        curFolds++;
        vector<int> nString(curString.begin() + m, curString.end());
        reset();
        for (int x : nString) add_char(x);
      }
    }
    ans.push_back(curFolds + numSuffFolds.back());
  }
  for (int i = 0; i < N; i++) {
    cout << ans[i] << " \n"[i + 1 == N];
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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() {
  using namespace std;
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int N;
  cin >> N;
  vector<int> A(N);
  for (auto& a : A) {
    cin >> a;
    a--;
  }
  vector<int> ans;
  ans.reserve(N);
  vector<int> curString;
  curString.reserve(N);
  vector<int> numSuffFolds;
  numSuffFolds.reserve(N + 1);
  vector<vector<int>> suffixPalindromes;
  suffixPalindromes.reserve(N + 1);
  auto reset = [&]() {
    curString.clear();
    numSuffFolds.clear();
    numSuffFolds.push_back(0);
    suffixPalindromes.clear();
    suffixPalindromes.emplace_back();
  };
  auto add_char = [&](int v) -> void {
    int r = int(curString.size()) + 1;
    vector<int> nSuffixPalindromes;
    nSuffixPalindromes.reserve(suffixPalindromes.back().size() + 1);
    if (!curString.empty() && curString.back() == v)
      nSuffixPalindromes.push_back(r - 1);
    for (int m : suffixPalindromes.back()) {
      if (2 * m - r >= 0 && v == curString[2 * m - r]) {
        nSuffixPalindromes.push_back(m);
      }
    }
    curString.push_back(v);
    suffixPalindromes.push_back(nSuffixPalindromes);
    numSuffFolds.push_back(nSuffixPalindromes.empty()
                               ? 0
                               : numSuffFolds[nSuffixPalindromes[0]] + 1);
  };
  int curFolds = 0;
  reset();
  for (int v : A) {
    add_char(v);
    int r = int(curString.size());
    if (suffixPalindromes.back().size() == 1) {
      int m = suffixPalindromes.back()[0];
      int l = m - (r - m);
      if (!suffixPalindromes[m].empty() && suffixPalindromes[m][0] == l) {
        curFolds += 2;
        curString.resize(l);
        numSuffFolds.resize(l + 1);
        suffixPalindromes.resize(l + 1);
      } else if (l == 0) {
        curFolds++;
        vector<int> nString(curString.begin() + m, curString.end());
        reset();
        for (int x : nString) add_char(x);
      }
    }
    ans.push_back(curFolds + numSuffFolds.back());
  }
  for (int i = 0; i < N; i++) {
    cout << ans[i] << " \n"[i + 1 == N];
  }
  return 0;
}

```