Codeforces-Round-#356-(Div.-2)-Bear-and-Finding-Criminals

Bear and Finding Criminals

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n cities in Bearland, numbered 1 through n. Cities are arranged in one long row. The distance between cities i and j is equal to |i - j|.

Limak is a police officer. He lives in a city a. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.

Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a citya. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.

You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.

Input

The first line of the input contains two integers n and a (1 ≤ a ≤ n ≤ 100) — the number of cities and the index of city where Limak lives.

The second line contains n integers t1, t2, ..., tn (0 ≤ ti ≤ 1). There are ti criminals in the i-th city.

Output

Print the number of criminals Limak will catch.

Examples

input

6 3
1 1 1 0 1 0

output

3

input

5 2
0 0 0 1 0

output

1

Note

In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.

Using the BCD gives Limak the following information:

• There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city.
• There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city.
• There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city.
• There are zero criminals for every greater distance.

So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.

In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.

      题意：告诉你n个城市上的罪犯个数，以及警察所在位置，警察能够知道距离他附近i单位的城市上是否有罪犯，但不知道他在哪个城市，要你求他最多能抓到的罪犯个数。 题目链接：Bear and Finding Criminals 解题思路：          分2种情况。 一.与位置p相距i单位的左边，右边城市都有罪犯，那么随便跑哪个城市都能抓到罪犯，且两个都能抓到。 二.与位置p相距i单位的左边（右边）没有城市，但是对称的右边（左边）有城市，且城市里面有罪犯，也就是有罪犯的城市唯一，那么这个罪犯就一定能被抓到。 代码：

//B. Bear and Finding Criminals
//题目链接：http://www.codeforces.com/problemset/problem/680/B
#include
#include
#include
#include
using namespace std;
int a[105];
int main()
{
    int n,p,i,ans;
    while(cin>>n>>p){
        for(i=0;i= n && p - i < 0)      break;   //已经扫描完毕
            if(p + i >= n && a[p-i]) ans++;   //右边距离i位置已经没有城市，但左边距离i位置有罪犯
            else if(p - i < 0 && a[p+i]) ans++; //左边距离i位置已经没有城市，但右边距离i位置有罪犯
            else if(p + i < n && p - i >= 0 && a[p-i] && a[p+i]) ans+=2; //左右两边距离自己i位置都有城市且都有罪犯
        }
        printf("%d\n",ans);
    }
    return 0;
}