codeforces 558B B. Amr and The Large Array(水题)

题目链接：

B. Amr and The Large Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

 

Amr has got a large array of size n. Amr doesn't like large arrays so he intends to make it smaller.

Amr doesn't care about anything in the array except the beauty of it. The beauty of the array is defined to be the maximum number of times that some number occurs in this array. He wants to choose the smallest subsegment of this array such that the beauty of it will be the same as the original array.

Help Amr by choosing the smallest subsegment possible.

 

Input

The first line contains one number n (1 ≤ n ≤ 105), the size of the array.

The second line contains n integers ai (1 ≤ ai ≤ 106), representing elements of the array.

 

Output

Output two integers l, r (1 ≤ l ≤ r ≤ n), the beginning and the end of the subsegment chosen respectively.

If there are several possible answers you may output any of them.

 

Examples

input

5
1 1 2 2 1

output

1 5

input

5
1 2 2 3 1

output

2 3

input

6
1 2 2 1 1 2

output

1 5

Note

A subsegment B of an array A from l to r is an array of size r - l + 1 where Bi = Al + i - 1 for all 1 ≤ i ≤ r - l + 1

 

题意：

 

使数组的魅力值保持相同,但数组的长度尽量小;

 

思路：

 

用vector的大小来看是否对应魅力值的那个数,而vector保存了这个数的各个位置,可以得到长度,一半比较一边更新就好;

 

AC代码：

/*
2014300227     558B - 9    GNU C++11    Accepted    61 ms    15728 KB
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
typedef long long ll;
const double PI=acos(-1.0);
int n,a[N];
vector<int>ve[10*N];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        ve[a[i]].push_back(i);
    }
    int ans=0,l=0,r=0;
    for(int i=1;i<=1e6;i++)
    {
        int g=ve[i].size();
        if(g){
        if(g>ve[ans].size())
        {
            int len=g;
            l=ve[i][0];
            r=ve[i][len-1];
            ans=i;
        }
        else if(g==ve[ans].size())
        {
            int s=g;
            if(ve[i][s-1]-ve[i][0]<r-l)
            {
                l=ve[i][0];
                r=ve[i][s-1];
                ans=i;
            }
        }
        }
    }
    cout<<l<<" "<<r<<endl;

    return 0;
}