/*
离散化后树状数组上二分。整体二分
时间复杂度n*logn*logn
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 2e5 + 5;

ll n, a[N], b[N], m, sum, R[N];
struct qry{
	ll gs, hei;
}Q[N];

ll tree[N];
void upd(ll x, ll k){
	for (; x <= n; x += (x & (-x)))
		tree[x] += k;
}
ll que(ll x){
	ll res = 0;
	for (; x; x -= (x & (-x)))
		res += tree[x];
	return res;
}

ll l, r, mid, rk;

int main(){
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin>>n;
	for (int i = 1; i <= n; ++i)
		cin>>Q[i].hei>>Q[i].gs, a[i] = b[i] = Q[i].hei;
	sort(a + 1, a + n + 1);
	m = unique(a + 1, a + n + 1) - (a + 1);
	for (int i = 1; i <= n; ++i)
		Q[i].hei = lower_bound(a + 1, a + m + 1, Q[i].hei) - a, R[Q[i].hei] = b[i];
	for (int i = 1; i <= n; ++i){
		sum += Q[i].gs, upd(Q[i].hei, Q[i].gs);
		l = 1, r = n, rk = 0;
		while (l <= r){//logn*logn 树状数组上二分。找到sum/2所在的位置。
			mid = (l + r) >> 1;
			if (que(mid) >= ((sum + 1) >> 1))	r = mid - 1, rk = mid;
			else	l = mid + 1;
		}
		cout<<R[rk]<<'\n';
	}

	return 0;
}