976 C. Nested Segments

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

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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10⁹) — the i-th segment.

Output

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

Examples

Input

Copy

5
1 10
2 9
3 9
2 3
2 9

Output

Copy

2 1

Input

Copy

3
1 5
2 6
6 20

Output

Copy

-1 -1
寻找是否存在一个段包含在另一个段里，输出段的编号

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
struct node
{
    int x,y,z;
    bool operator<(const node &a)
    {
        return a.x==x?a.y<y:a.x>x;
    }
}e[300006];
int n;
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&e[i].x,&e[i].y);
        e[i].z=i;
    }
    sort(e,e+n);
    for(int i=0;i<n-1;i++)
    {
        if(e[i].x<=e[i+1].x && e[i].y>=e[i+1].y)
        {
            printf("%d %d\n",e[i+1].z+1,e[i].z+1);
            return 0;
        }
    }
    printf("-1 -1\n");
    return 0;
}