Problem: For a given set of two-dimensional points S, let's denote its extension E(S) as the result of the following algorithm:

Create another set of two-dimensional points R, which is initially equal to S. Then, while there exist four numbers x_1, y_1, x_2 and y_2 such that (x_1, y_1) ∈ R, (x_1, y_2) ∈ R, (x_2, y_1) ∈ R and (x_2, y_2) ∉ R, add (x_2, y_2) to R. When it is impossible to find such four integers, let R be the result of the algorithm.

Now for the problem itself. You are given a set of two-dimensional points S, which is initially empty. You have to process two types of queries: add some point to S, or remove some point from it. After each query you have to compute the size of E(S).

Input

The first line contains one integer q (1 ≤ q ≤ 3 ⋅ 10^5) — the number of queries.

Then q lines follow, each containing two integers x_i, y_i (1 ≤ x_i, y_i ≤ 3 ⋅ 10^5), denoting i-th query as follows: if (x_i, y_i) ∈ S, erase it from S, otherwise insert (x_i, y_i) into S.

Output

Print q integers. i-th integer should be equal to the size of E(S) after processing first i queries.

Example

Input


7
1 1
1 2
2 1
2 2
1 2
1 3
2 1


Output


1 2 4 4 4 6 3 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
template <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  register bool f = 0;
  while (!isdigit(c)) f ^= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  if (f) x = -x;
}
template <class T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 3e5 + 10;
int n;
long long ans, out[N];
std::map<std::pair<int, int>, int> map;
struct info {
  int x, y, l, r;
};
std::vector<info> q;
struct node {
  int u, l, r, mid;
  std::vector<std::pair<int, int>> vet;
} p[N << 2];
struct atom {
  int f, a, b;
  long long s() { return (long long)a * b; }
} f[N << 1];
int find(int x) { return f[x].f == x ? x : find(f[x].f); }
void merge(int u, int v, std::vector<std::pair<int, atom>> &rub) {
  u = find(u), v = find(v);
  if (u == v) return;
  if (f[u].a + f[u].b > f[v].a + f[v].b) std::swap(u, v);
  ans -= f[u].s(), rub.push_back(std::make_pair(u, f[u]));
  ans -= f[v].s(), rub.push_back(std::make_pair(v, f[v]));
  f[u].f = v, f[v].a += f[u].a, f[v].b += f[u].b, ans += f[v].s();
}
void build(int u, int l, int r) {
  p[u].l = l, p[u].r = r, p[u].mid = (l + r) >> 1;
  if (l == r) return;
  build(u << 1, l, p[u].mid), build(u << 1 | 1, p[u].mid + 1, r);
}
void insert(int u, int l, int r, std::pair<int, int> v) {
  if (p[u].l == l && p[u].r == r) {
    p[u].vet.push_back(v);
    return;
  }
  if (r <= p[u].mid)
    insert(u << 1, l, r, v);
  else if (l > p[u].mid)
    insert(u << 1 | 1, l, r, v);
  else
    insert(u << 1, l, p[u].mid, v), insert(u << 1 | 1, p[u].mid + 1, r, v);
}
void dfs(int u) {
  std::vector<std::pair<int, atom>> rub;
  long long rubans = ans;
  for (const auto &x : p[u].vet) merge(x.first, x.second + N, rub);
  if (p[u].l == p[u].r)
    out[p[u].l] = ans;
  else
    dfs(u << 1), dfs(u << 1 | 1);
  for (const auto &x : (std::reverse(rub.begin(), rub.end()), rub))
    f[x.first] = x.second;
  ans = rubans;
}
int main() {
  for (int i = 0; i < N; i++) f[i] = {i, 1, 0}, f[i + N] = {i + N, 0, 1};
  read(n);
  for (int x, y, i = 1; i <= n; i++) {
    read(x), read(y);
    auto it = std::make_pair(x, y);
    map[it] ? (q.push_back({x, y, map[it], i - 1}), map[it] = 0)
            : (map[it] = i);
  }
  for (const auto &it : map)
    if (it.second) q.push_back({it.first.first, it.first.second, it.second, n});
  build(1, 1, n);
  for (const auto &it : q) insert(1, it.l, it.r, std::make_pair(it.x, it.y));
  dfs(1);
  for (int i = 1; i <= n; i++) print(out[i]), putchar(' ');
}