#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, a[N], m, x, i, Z;
unordered_map<int, int> mp[N];
vector<int> b[N], bb[N];
int dfs(int l, int r) {
  if (mp[r].count(l)) return mp[r][l];
  int& ans = mp[r][l] = 0;
  for (int y : b[r])
    if (r - y * 2 + 1 >= l) ans = max(ans, dfs(l, r - y) + 1);
  for (int y : bb[l])
    if (l + y * 2 - 1 <= r) ans = max(ans, dfs(l + y, r) + 1);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (; n--;) {
    scanf("%d", &x);
    a[m + 1] = x;
    if (a[m] == x) b[m + 1] = {1};
    for (int y : b[m])
      if (a[m - y * 2] == x) b[m + 1].push_back(y + 1);
    ++m;
    i = 0;
    for (int y : b[m])
      if (binary_search(b[m - y].begin(), b[m - y].end(), y)) i = y;
    for (int y : b[m]) bb[m - y * 2 + 1].push_back(y);
    if (i)
      for (i *= 2, ++Z; i--;) {
        for (int y : b[m]) bb[m - y * 2 + 1].pop_back();
        b[m].clear(), mp[m].clear(), --m;
      }
    else
      dfs(1, m), assert(mp[m].size() <= 19);
    printf("%d ", Z * 2 + mp[m][1]);
  }
  puts("");
}

### Prompt

Please create a solution in cpp 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;
const int N = 1e5 + 5;
int n, a[N], m, x, i, Z;
unordered_map<int, int> mp[N];
vector<int> b[N], bb[N];
int dfs(int l, int r) {
  if (mp[r].count(l)) return mp[r][l];
  int& ans = mp[r][l] = 0;
  for (int y : b[r])
    if (r - y * 2 + 1 >= l) ans = max(ans, dfs(l, r - y) + 1);
  for (int y : bb[l])
    if (l + y * 2 - 1 <= r) ans = max(ans, dfs(l + y, r) + 1);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (; n--;) {
    scanf("%d", &x);
    a[m + 1] = x;
    if (a[m] == x) b[m + 1] = {1};
    for (int y : b[m])
      if (a[m - y * 2] == x) b[m + 1].push_back(y + 1);
    ++m;
    i = 0;
    for (int y : b[m])
      if (binary_search(b[m - y].begin(), b[m - y].end(), y)) i = y;
    for (int y : b[m]) bb[m - y * 2 + 1].push_back(y);
    if (i)
      for (i *= 2, ++Z; i--;) {
        for (int y : b[m]) bb[m - y * 2 + 1].pop_back();
        b[m].clear(), mp[m].clear(), --m;
      }
    else
      dfs(1, m), assert(mp[m].size() <= 19);
    printf("%d ", Z * 2 + mp[m][1]);
  }
  puts("");
}

```