Educational Codeforces Round 38 (Rated for Div. 2) 【B】【双指针】

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本文介绍了一种解决电视游戏“RunForYourPrize”的算法。游戏中，两名玩家需合作收集位于直线上的奖品，目标是在最短时间内完成任务。算法通过双指针技巧，在O(n)时间内找到最优解。

You and your friend are participating in a TV show "Run For Your Prize".

At the start of the show n prizes are located on a straight line. i-th prize is located at position ai. Positions of all prizes are distinct. You start at position 1, your friend — at position 106 (and there is no prize in any of these two positions). You have to work as a team and collect all prizes in minimum possible time, in any order.

You know that it takes exactly 1 second to move from position x to position x + 1 or x - 1, both for you and your friend. You also have trained enough to instantly pick up any prize, if its position is equal to your current position (and the same is true for your friend). Carrying prizes does not affect your speed (or your friend's speed) at all.

Now you may discuss your strategy with your friend and decide who will pick up each prize. Remember that every prize must be picked up, either by you or by your friend.

What is the minimum number of seconds it will take to pick up all the prizes?

Input

The first line contains one integer n (1 ≤ n ≤ 105) — the number of prizes.

The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 106 - 1) — the positions of the prizes. No two prizes are located at the same position. Positions are given in ascending order.

Output

Print one integer — the minimum number of seconds it will take to collect all prizes.

Examples

input

Copy
3
2 3 9

output
8

input

Copy
2
2 999995

output
5

Note

In the first example you take all the prizes: take the first at 1, the second at 2 and the third at 8.

In the second example you take the first prize in 1 second and your friend takes the other in 5 seconds, you do this simultaneously, so the total time is 5.

#include<bits/stdc++.h>
using namespace std;
#define MAX 1000000
#define N 100005
int a[N];
int main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    int x = 0, y = n - 1, ans = 0;
    for(int i = 2, j = MAX - 1; i <= j && x <= y; i++, j--)
    {
        ans++;
        if(i == a[x])
            x++;
        if(j == a[y])
            y--;
    }
    cout << ans << endl;
    return 0;
}