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>
using namespace std;
int q;
map<int, int> mp[600005];
struct edg {
  int u, v;
  edg() {}
  edg(int u, int v) : u(u), v(v) {}
};
vector<edg> G[1048576];
void Add(int i, int l, int r, int a, int b, edg x) {
  if (r < a || b < l) return;
  if (a <= l && r <= b) {
    G[i].push_back(x);
    return;
  }
  Add(i << 1, l, (l + r) >> 1, a, b, x),
      Add(i << 1 | 1, ((l + r) >> 1) + 1, r, a, b, x);
}
long long Ans[600005];
int fa[600005], sz[600005], lb[600005], rb[600005];
int ff(int x) { return fa[x] ? ff(fa[x]) : x; }
int stk[600005], tp;
long long ans;
void DFS(int i, int l, int r) {
  int ot = tp;
  for (auto x : G[i]) {
    int u = ff(x.u), v = ff(3e5 + x.v);
    if (u == v) continue;
    if (sz[u] < sz[v]) swap(u, v);
    ans += (long long)lb[u] * rb[v] + (long long)rb[u] * lb[v];
    fa[v] = u, lb[u] += lb[v], rb[u] += rb[v], sz[u] += sz[v];
    stk[++tp] = v;
  }
  if (l == r)
    Ans[l] = ans;
  else
    DFS(i << 1, l, (l + r) >> 1), DFS(i << 1 | 1, ((l + r) >> 1) + 1, r);
  while (tp != ot) {
    int u = stk[tp--], v = fa[u];
    lb[v] -= lb[u], rb[v] -= rb[u], sz[v] -= sz[u];
    fa[u] = 0;
    ans -= (long long)lb[u] * rb[v] + (long long)rb[u] * lb[v];
  }
}
int main() {
  scanf("%d", &q);
  for (int i = 1; i <= (6e5); ++i) (i <= 3e5 ? lb : rb)[i] = sz[i] = 1;
  for (int i = 1; i <= (q); ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (mp[u][v])
      Add(1, 1, q, mp[u][v], i - 1, edg(u, v)), mp[u][v] = 0;
    else
      mp[u][v] = i;
  }
  for (int i = 1; i <= (3e5); ++i)
    for (auto x : mp[i])
      if (x.second) Add(1, 1, q, x.second, q, edg(i, x.first));
  DFS(1, 1, q);
  for (int i = 1; i <= (q); ++i) printf("%lld ", Ans[i]);
  return 0;
}