A. Reconnaissance 2

A. Reconnaissance 2

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouringsoldiers, whose heights difference is minimal, i.e. |ai?-?aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.

Input

The first line contains integer n (2?≤?n?≤?100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle —n space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?1000). The soldier heights are given in clockwise or counterclockwise direction.

Output

Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.

Sample test(s)

input

5
10 12 13 15 10

output

5 1

input

4
10 20 30 40

output

1 2

/* ***********************************************
Author        :
Created Time  :2015/6/14 12:18:25
File Name     :7.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1<<30
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int a[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
	int n;
	int x,y;
	while(cin>>n){
		for(int i=1;i<=n;i++)
			cin>>a[i];
		a[n+1]=a[1];
		int Min=INF;
		for(int i=2;i<=n+1;i++){
				if(abs(a[i]-a[i-1])<Min){
						Min=abs(a[i]-a[i-1]);
						x=i;y=i-1;
						if(x>n)x=1;
					
				}
		}
		cout<<y<<" "<<x<<endl;
	}
    return 0;
}