Suppose you are given a sequence S of k pairs of integers (a_1, b_1), (a_2, b_2), ..., (a_k, b_k).

You can perform the following operations on it:

  1. Choose some position i and increase a_i by 1. That can be performed only if there exists at least one such position j that i ≠ j and a_i = a_j. The cost of this operation is b_i; 
  2. Choose some position i and decrease a_i by 1. That can be performed only if there exists at least one such position j that a_i = a_j + 1. The cost of this operation is -b_i. 



Each operation can be performed arbitrary number of times (possibly zero).

Let f(S) be minimum possible x such that there exists a sequence of operations with total cost x, after which all a_i from S are pairwise distinct. 

Now for the task itself ...

You are given a sequence P consisting of n pairs of integers (a_1, b_1), (a_2, b_2), ..., (a_n, b_n). All b_i are pairwise distinct. Let P_i be the sequence consisting of the first i pairs of P. Your task is to calculate the values of f(P_1), f(P_2), ..., f(P_n).

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of pairs in sequence P.

Next n lines contain the elements of P: i-th of the next n lines contains two integers a_i and b_i (1 ≤ a_i ≤ 2 ⋅ 10^5, 1 ≤ b_i ≤ n). It is guaranteed that all values of b_i are pairwise distinct.

Output

Print n integers — the i-th number should be equal to f(P_i).

Examples

Input

5
1 1
3 3
5 5
4 2
2 4


Output

0
0
0
-5
-16


Input

4
2 4
2 3
2 2
1 1


Output

0
3
7
1
#include <bits/stdc++.h>
const int N = 400005;
int root[N], size[N * 10], ls[N * 10], rs[N * 10], cnt, f[N], r[N], n, a, b;
long long sum[N * 10], ans;
void pushup(int x) {
  size[x] = size[ls[x]] + size[rs[x]];
  sum[x] = sum[ls[x]] + sum[rs[x]];
}
void ins(int &k, int l, int r, int x) {
  if (!k) k = ++cnt;
  if (l == r) {
    size[k] = 1;
    sum[k] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    ins(ls[k], l, mid, x);
  else
    ins(rs[k], mid + 1, r, x);
  pushup(k);
}
long long c(int x) { return size[rs[x]] * sum[ls[x]]; }
int merge(int x, int y) {
  if (!x) return y;
  if (!y) return x;
  ans = ans - c(x) - c(y);
  ls[x] = merge(ls[x], ls[y]);
  rs[x] = merge(rs[x], rs[y]);
  pushup(x), ans = ans + c(x);
  return x;
}
int find(int x) {
  if (f[x] == x) return x;
  f[x] = find(f[x]);
  return f[x];
}
void solve(int x, int y) {
  int fx = find(x), fy = find(y);
  ans = ans + sum[root[y]] * (fx - fy);
  root[fx] = merge(root[fx], root[fy]);
  f[fy] = fx, r[fx] = r[fy];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 400000; i++) f[i] = i;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a, &b);
    int fa = find(a), p = r[fa] ? r[fa] + 1 : a;
    r[p] = p;
    ans = ans - 1ll * a * b;
    ins(root[p], 1, n, b);
    ans = ans + 1ll * p * b;
    if (r[p - 1]) solve(p - 1, p);
    if (r[p + 1]) solve(p, p + 1);
    printf("%lld\n", ans);
  }
}
