#include <bits/stdc++.h>
using namespace std;
int n, A[100005];
int xyx, q[100005], c[100005], p[100005];
vector<int> v[100005];
int pos;
void get_v(int pos) {
  v[pos].clear();
  if (pos == 1) return;
  if (A[pos] == A[pos - 1]) v[pos].push_back(2);
  for (auto x : v[pos - 1])
    if (A[pos] == A[pos - x - 1]) v[pos].push_back(x + 2);
}
bool Check(int L, int R) {
  for (auto x : v[R])
    if (x == R - L + 1) return 1;
  return 0;
}
int Query() {
  int res = xyx + c[pos];
  int cur = pos;
  while (p[cur] && q[pos] <= cur - p[cur] / 2) {
    cur -= p[cur] / 2;
    ++res;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  q[0] = 1;
  for (int i = 1; i <= n; i++) {
    ++pos;
    scanf("%d", &A[pos]);
    get_v(pos);
    int x;
    if (v[pos].size() &&
        (x = v[pos][0], Check(pos - x / 2 - x + 1, pos - x / 2))) {
      pos -= x;
      xyx += 2;
    } else {
      p[pos] = v[pos].size() ? v[pos][0] : 0;
      q[pos] = q[pos - 1];
      c[pos] = c[pos - 1];
      if (Check(q[pos], pos)) {
        q[pos] += (pos - q[pos] + 1) / 2;
        c[pos]++;
      }
    }
    printf("%d%c", Query(), i == n ? 10 : 32);
  }
}

### 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;
int n, A[100005];
int xyx, q[100005], c[100005], p[100005];
vector<int> v[100005];
int pos;
void get_v(int pos) {
  v[pos].clear();
  if (pos == 1) return;
  if (A[pos] == A[pos - 1]) v[pos].push_back(2);
  for (auto x : v[pos - 1])
    if (A[pos] == A[pos - x - 1]) v[pos].push_back(x + 2);
}
bool Check(int L, int R) {
  for (auto x : v[R])
    if (x == R - L + 1) return 1;
  return 0;
}
int Query() {
  int res = xyx + c[pos];
  int cur = pos;
  while (p[cur] && q[pos] <= cur - p[cur] / 2) {
    cur -= p[cur] / 2;
    ++res;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  q[0] = 1;
  for (int i = 1; i <= n; i++) {
    ++pos;
    scanf("%d", &A[pos]);
    get_v(pos);
    int x;
    if (v[pos].size() &&
        (x = v[pos][0], Check(pos - x / 2 - x + 1, pos - x / 2))) {
      pos -= x;
      xyx += 2;
    } else {
      p[pos] = v[pos].size() ? v[pos][0] : 0;
      q[pos] = q[pos - 1];
      c[pos] = c[pos - 1];
      if (Check(q[pos], pos)) {
        q[pos] += (pos - q[pos] + 1) / 2;
        c[pos]++;
      }
    }
    printf("%d%c", Query(), i == n ? 10 : 32);
  }
}

```