#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
const long long MOD2 = (long long)1000000007 * (long long)1000000007;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
vector<int> sufPaliLen[(100035)];
int a[(100035)], p[(100035)], q[(100035)], pos, c[(100035)], xyx;
void add(int pos) {
  sufPaliLen[pos].clear();
  if (a[pos] == a[pos - 1]) sufPaliLen[pos].push_back(2);
  for (int x : sufPaliLen[pos - 1]) {
    if (a[pos - 1 - x] == a[pos]) sufPaliLen[pos].push_back(x + 2);
  }
}
bool ckEvenPali(int l, int r) {
  for (int z : sufPaliLen[r]) {
    if (z == r - l + 1) return 1;
  }
  return 0;
}
int qy() {
  int ans = xyx + c[pos];
  int cur = pos;
  while (p[cur] && q[pos] <= cur - p[cur] / 2) {
    cur -= p[cur] / 2;
    ans++;
  }
  return ans;
}
void fmain(int tid) {
  scanf("%d", &n);
  q[0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    pos++;
    scanf("%d", a + pos);
    bool fg = 0;
    add(pos);
    if (!sufPaliLen[pos].empty()) {
      int len = sufPaliLen[pos][0];
      if (ckEvenPali(pos - len / 2 - len + 1, pos - len / 2)) {
        pos -= len;
        xyx += 2;
        fg = 1;
      }
    }
    if (!fg) {
      p[pos] = sufPaliLen[pos].empty() ? 0 : sufPaliLen[pos][0];
      q[pos] = q[pos - 1];
      c[pos] = c[pos - 1];
      if (ckEvenPali(q[pos], pos)) {
        q[pos] += (pos - q[pos] + 1) / 2;
        c[pos]++;
      }
    }
    printf("%d ", qy());
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}

### Prompt

Generate a cpp solution 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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
const long long MOD2 = (long long)1000000007 * (long long)1000000007;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
vector<int> sufPaliLen[(100035)];
int a[(100035)], p[(100035)], q[(100035)], pos, c[(100035)], xyx;
void add(int pos) {
  sufPaliLen[pos].clear();
  if (a[pos] == a[pos - 1]) sufPaliLen[pos].push_back(2);
  for (int x : sufPaliLen[pos - 1]) {
    if (a[pos - 1 - x] == a[pos]) sufPaliLen[pos].push_back(x + 2);
  }
}
bool ckEvenPali(int l, int r) {
  for (int z : sufPaliLen[r]) {
    if (z == r - l + 1) return 1;
  }
  return 0;
}
int qy() {
  int ans = xyx + c[pos];
  int cur = pos;
  while (p[cur] && q[pos] <= cur - p[cur] / 2) {
    cur -= p[cur] / 2;
    ans++;
  }
  return ans;
}
void fmain(int tid) {
  scanf("%d", &n);
  q[0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    pos++;
    scanf("%d", a + pos);
    bool fg = 0;
    add(pos);
    if (!sufPaliLen[pos].empty()) {
      int len = sufPaliLen[pos][0];
      if (ckEvenPali(pos - len / 2 - len + 1, pos - len / 2)) {
        pos -= len;
        xyx += 2;
        fg = 1;
      }
    }
    if (!fg) {
      p[pos] = sufPaliLen[pos].empty() ? 0 : sufPaliLen[pos][0];
      q[pos] = q[pos - 1];
      c[pos] = c[pos - 1];
      if (ckEvenPali(q[pos], pos)) {
        q[pos] += (pos - q[pos] + 1) / 2;
        c[pos]++;
      }
    }
    printf("%d ", qy());
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}

```