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本文介绍了一个基于纸面坐标系统的有趣游戏，玩家通过添加、删除坐标点及查询特定坐标点右侧上方最近的点来互动。文章提供了完整的算法实现，包括如何高效处理大量请求。

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Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:

add x y — on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
remove x y — on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
find x y — on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.

Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!

Input

The first input line contains number n (1 ≤ n ≤ 2·105) — amount of requests. Then there follow n lines — descriptions of the requests. add x y describes the request to add a point, remove x y — the request to erase a point, find x y — the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.

Output

For each request of type find x y output in a separate line the answer to it — coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.

Examples

Input

7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0

Output

1 1
3 4
1 1

Input

13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4

Output

7 7
-1
5 5

#include<bits/stdc++.h>
using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
const int MAX = 2e6 + 7;
pair <int, int> p[MAX], X[MAX], ans;
int a[MAX << 2];
void pushup(int rt){
    a[rt] = max(a[rt << 1], a[rt << 1 | 1]);
}
bool cmp(pair <int, int> a, pair <int, int> b){
    if(a.first == b.first) return a.second < b.second;
    return a.first < b.first;
}
void update(int rt, int l, int r, int pos, int v, int op){
    if(l == r){
        a[rt] = op ? v : -1;
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) update(lson, pos, v, op);
    else update(rson, pos, v, op);
    pushup(rt);
}
void query(int rt, int l, int r, pair <int, int> v){
    if(a[rt] <= v.second || X[r].first <= v.first) return;
    if(l == r) ans = X[l];
    int mid = (l + r) >> 1;
    query(lson, v);
    if(ans.first == -1) query(rson, v);
}
int flag[MAX];
int main(){
    memset(a, -1, sizeof a);
    char s[5];
    int n, tot = 0;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%s%d%d", s, &p[i].first, &p[i].second);
        if(s[0] == 'a') X[tot++] = p[i], flag[i] = 1;
        else if(s[0] == 'f') X[tot++] = p[i], flag[i] = 2;
        else flag[i] = 0;
    }
    sort(X, X + tot, cmp);
    tot = unique(X, X + tot) - X;
    for(int i = 0; i < n; i++){
        int pos = lower_bound(X, X + tot, p[i]) - X;
        if(flag[i] != 2) update(1, 0, tot - 1, pos, p[i].second, flag[i]);
        else{
            ans = make_pair(-1, -1);
            query(1, 0, tot - 1, p[i]);
            if(ans.first == -1) puts("-1");
            else printf("%d %d\n", ans.first, ans.second);
        }
    }
    return 0;
}