#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, top;
vector<int> f[N];
int a[N], s[N];
void ins(int x) {
  f[x].clear();
  if (a[x] == a[x - 1]) f[x].push_back(1);
  for (auto i : f[x - 1])
    if (a[x - 2 * i - 1] == a[x]) f[x].push_back(i + 1);
}
int check(int x) {
  if (f[x].empty()) return 0;
  int d = f[x].front();
  return binary_search(f[x - d].begin(), f[x - d].end(), d) ? d : 0;
}
int suf[N], p[N], c[N];
void push(int x) {
  suf[x] = (f[x].empty() ? 0 : f[x].front());
  p[x] = p[x - 1];
  c[x] = c[x - 1];
  int tmp = x - p[x];
  if (!(tmp & 1) && binary_search(f[x].begin(), f[x].end(), tmp >>= 1))
    p[x] += tmp, c[x]++;
}
int query(int x) {
  int res = c[x], l = p[x];
  for (int i = x; suf[i] && l < i - suf[i]; i -= suf[i]) res++;
  return res;
}
int z;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    top++;
    scanf("%d", &a[top]);
    ins(top);
    int tmp;
    if ((tmp = check(top)))
      top -= 2 * tmp, z++;
    else
      push(top);
    printf("%d ", 2 * z + query(top));
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 100010;
int n, top;
vector<int> f[N];
int a[N], s[N];
void ins(int x) {
  f[x].clear();
  if (a[x] == a[x - 1]) f[x].push_back(1);
  for (auto i : f[x - 1])
    if (a[x - 2 * i - 1] == a[x]) f[x].push_back(i + 1);
}
int check(int x) {
  if (f[x].empty()) return 0;
  int d = f[x].front();
  return binary_search(f[x - d].begin(), f[x - d].end(), d) ? d : 0;
}
int suf[N], p[N], c[N];
void push(int x) {
  suf[x] = (f[x].empty() ? 0 : f[x].front());
  p[x] = p[x - 1];
  c[x] = c[x - 1];
  int tmp = x - p[x];
  if (!(tmp & 1) && binary_search(f[x].begin(), f[x].end(), tmp >>= 1))
    p[x] += tmp, c[x]++;
}
int query(int x) {
  int res = c[x], l = p[x];
  for (int i = x; suf[i] && l < i - suf[i]; i -= suf[i]) res++;
  return res;
}
int z;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    top++;
    scanf("%d", &a[top]);
    ins(top);
    int tmp;
    if ((tmp = check(top)))
      top -= 2 * tmp, z++;
    else
      push(top);
    printf("%d ", 2 * z + query(top));
  }
  return 0;
}

```