#include <bits/stdc++.h>
const int MAXN = 1e5 + 5;
int a[MAXN], n;
std::vector<int> v[MAXN];
int q[MAXN], c[MAXN];
inline bool pd(int l, int r) {
  for (auto x : v[r])
    if (x == r - l + 1) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  int now = 0;
  q[0] = 1;
  int xyx = 0;
  for (int ccc = 1; ccc <= n; ++ccc) {
    ++now;
    scanf("%d", a + now);
    v[now].clear();
    if (now != 1) {
      if (a[now] == a[now - 1]) v[now].push_back(2);
      for (auto x : v[now - 1])
        if (a[now - x - 1] == a[now]) v[now].push_back(x + 2);
    }
    if (!v[now].empty() &&
        pd(now - v[now][0] / 2 - v[now][0] + 1, now - v[now][0] / 2)) {
      now -= v[now][0];
      xyx += 2;
    } else {
      q[now] = q[now - 1];
      c[now] = c[now - 1];
      if (pd(q[now], now)) {
        q[now] += (now - q[now] + 1) / 2;
        c[now]++;
      }
    }
    int ans = xyx + c[now], p = now;
    while (!v[p].empty() && p - v[p][0] / 2 >= q[now]) {
      p -= v[p][0] / 2;
      ans++;
    }
    printf("%d ", ans);
  }
  puts("");
  return 0;
}

### Prompt

Create a solution in CPP for 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>
const int MAXN = 1e5 + 5;
int a[MAXN], n;
std::vector<int> v[MAXN];
int q[MAXN], c[MAXN];
inline bool pd(int l, int r) {
  for (auto x : v[r])
    if (x == r - l + 1) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  int now = 0;
  q[0] = 1;
  int xyx = 0;
  for (int ccc = 1; ccc <= n; ++ccc) {
    ++now;
    scanf("%d", a + now);
    v[now].clear();
    if (now != 1) {
      if (a[now] == a[now - 1]) v[now].push_back(2);
      for (auto x : v[now - 1])
        if (a[now - x - 1] == a[now]) v[now].push_back(x + 2);
    }
    if (!v[now].empty() &&
        pd(now - v[now][0] / 2 - v[now][0] + 1, now - v[now][0] / 2)) {
      now -= v[now][0];
      xyx += 2;
    } else {
      q[now] = q[now - 1];
      c[now] = c[now - 1];
      if (pd(q[now], now)) {
        q[now] += (now - q[now] + 1) / 2;
        c[now]++;
      }
    }
    int ans = xyx + c[now], p = now;
    while (!v[p].empty() && p - v[p][0] / 2 >= q[now]) {
      p -= v[p][0] / 2;
      ans++;
    }
    printf("%d ", ans);
  }
  puts("");
  return 0;
}

```