题目链接:

http://codeforces.com/contest/976/problem/C


题目:

You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.

Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.

Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Input

The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.

Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.

Output

Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Examples

input

5
1 10
2 9
3 9
2 3
2 9

output

2 1

input

3
1 5
2 6
6 20

output

-1 -1

Note

In the first example the following pairs are considered correct:

(2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
(3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
(5, 2), (2, 5) — match exactly.


题意:

  给你n条线段,问你是否存在两条线段,使一条被另一条包含,如果有,先输出被包含的,然后再输出另一个,如果没有就输出两个$-1$。


思路:

  按照左端点升序,左端点相同时按照右端点降序排列,然后从左往右扫,因为左端点会严格递增,那么如果当前点的右端点小于等于前一个点的右端点,那么这条线段一定被前一条线段所包含。


实现:

#include <bits/stdc++.h>
const int maxn = int(3e5) + 7;
struct Node {
    int l, r, index;
    bool operator < (const Node &tmp) const {
        return l == tmp.l ? r > tmp.r : l < tmp.l;
    }
} node[maxn];
int main() {
//    freopen("in.txt", "r", stdin);
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d%d", &node[i].l, &node[i].r), node[i].index = i + 1;
    std::sort(node, node + n);
    for (int i = 1; i < n; i++) if (node[i].r <= node[i - 1].r) return 0 * printf("%d %d\n", node[i].index, node[i - 1].index);
    return 0 * puts("-1 -1");
}
最后修改:2018 年 05 月 01 日 01 : 08 AM
谢谢老板!