CodeForces 558B

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 a_i (1 ≤ a_i ≤ 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.

Sample Input

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


题意：一个整数序列的beauty值为序列中出现次数最多的整数的出现次数。求与给定整数序列的beauty值相同的该序列的最短子区间，只输出区间的始、末位置。

思路：1.先求出原序列的beauty值；2.然后找出出现次数与beauty值相同的元素；3.然后再找出元素的始末位置，以计算区间长度；4.最后择一个最小的区间就是答案。
     按上面思路，如果将数据按一般的方式存储来实现代码，会TLE。应将元素用结构体存储，以在3、4过程中节省时间。

# include<iostream> # include<cstdio> # include<map> # include<set> # include<vector> # include<cstring> using namespace std; struct node { int l,r,v; bool operator < (const node &a) const { return v<a.v; } }; map<node,int>m; int main() { int n,i,j; while(~scanf("%d",&n)) { m.clear(); node t; map<node,int>::iterator it; for(i=1;i<=n;++i){ scanf("%d",&t.v); it=m.find(t); if(it==m.end()){ t.l=t.r=i; ++m[t]; }else{ node tt=it->first; int res=it->second; m.erase(it->first); tt.r=i; m[tt]=res+1; } } int b=0; for(it=m.begin();it!=m.end();++it) b=max(b,it->second); //cout<<b<<endl; int ansl=1,ansr=n; for(it=m.begin();it!=m.end();++it){ if(it->second==b){ node tt=it->first; if(tt.r-tt.l<ansr-ansl) ansl=tt.l,ansr=tt.r; } } printf("%d %d\n",ansl,ansr); } return 0; }