Description：

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：

找到出现次数最多的数的区间，如果有多个输出区间最小的那一个

水题直接上代码

AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
#define maxn 1000010
int l[1000010],r[1000010],cnt[1000010];
int n,m;
int i,j,k,a,b,x;
int main()
{
    while(~scanf("%d",&n))
    {
        int ans=INF;
        int maxx=0;
        for(int i=0; i<maxn; i++)
        {
            l[i]=INF;
            r[i]=-INF;
            cnt[i]=0;
        }
        for(i=0; i<n; i++)
        {
            scanf("%d",&x);
            l[x]=min(l[x],i);//x点出现的最左点
            r[x]=max(r[x],i);//x点出现的最右点
            cnt[x]++;
        }
        for (i=0; i<maxn; i++)
            maxx=max(maxx,cnt[i]);
        for (i=0; i<maxn; i++)
            if(maxx==cnt[i]&&ans>r[i]-l[i]+1)
            {
                ans=r[i]-l[i]+1;
                j=i;//选择最小区间
            }
        printf("%d %d\n",l[j]+1,r[j]+1);
    }
    return 0;
}