Allen wants to enter a fan zone that occupies a round square and has nn entrances.
There already is a queue of aiai people in front of the ii-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
- Initially he stands in the end of the queue in front of the first entrance.
- Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer nn (2≤n≤1052≤n≤105) — the number of entrances.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤ai≤109) — the number of people in queues. These numbers do not include Allen.
Output
Print a single integer — the number of entrance that Allen will use.
Examples
input
4
2 3 2 0
output
3
input
2
10 10
output
1
input
6 5 2 6 5 7 4
output
6
time limit per test
1 second
memory limit per test
256 megabytes
Note
In the first example the number of people (not including Allen) changes as follows: [2,3,2,0]→[1,2,1,0]→[0,1,0,0][2,3,2,0]→[1,2,1,0]→[0,1,0,0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows:[10,10]→[9,9]→[8,8]→[7,7]→[6,6]→[5,5]→[4,4]→[3,3]→[2,2]→[1,1]→[0,0][10,10]→[9,9]→[8,8]→[7,7]→[6,6]→[5,5]→[4,4]→[3,3]→[2,2]→[1,1]→[0,0].
In the third example the number of people (not including Allen) changes as follows:[5,2,6,5,7,4]→[4,1,5,4,6,3]→[3,0,4,3,5,2]→[2,0,3,2,4,1]→[1,0,2,1,3,0]→[0,0,1,0,2,0]
代码如下：
#include<bits/stdc++.h>
const int maxn=1e5+10;
using namespace std;
int arr[maxn];
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
int lcount=0;
int index=0;
while((arr[index]-lcount)>0)
{
if(index==n-1)
index=0;
else
index++;
lcount++;
}
cout<<index+1<<endl;
}
代码2（思想差不多）
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int a[maxn];
int main()
{
int i,n;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i];
int k=0;
for(k=0;;k++)
{
for(i=1;i<=n;i++)
{
if(a[i]-n*k-i<0)
{
cout<<i<<endl;
return 0;
}
}
}
return 0;
}