Bear and Finding Criminals （模拟）

                                      Bear and Finding Criminals 

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 city a. 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.

大致题意：一个警察住在某个城市，他要抓各个城市的小偷，现在用一个BCD可以知道那个城市里一定有小偷。

                  (一定能确定该城市有小偷的几种情况：

                     1.警察所住城市有小偷，则一定能检测到

                     2.警察所住城市的左面和右面位置若都不为0，则说明两座城市都有小偷

                     3.警察所在城市的一边检测到有小偷，但在另一边已经没有城市了，则说明该城市一定有小偷)

第一次做的时候，从警察所在城市向两边扩展，判断两边所存数组的值是否为0，结果测试到第15组数据是出错，后来才发现，忽略了数组之外的值，他们的值是随机的。

错误代码

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

const int MAXN = 1000;
int t[MAXN];

int main()
{
	int n, a;
	while(scanf("%d%d",&n, &a)!=EOF){
		int cnt = 0;
		memset(t, 0, sizeof(t));
		for(int i =1; i <= n; i++)
			cin >> t[i];
		if(t[a]) cnt++;
		cout << cnt <<endl;
		for(int i = 1; i <= n; i++){
			if(t[a-i] >= 1&&t[a+i] == 1) {  //******错误点 
				cnt+=2;
				cout << cnt;
			}
			else if(a-i <= 0&&a+i <= n){
				if(t[a+i]) cnt++;
				cout << cnt ;
			}
			else if(a-i >= 1&&a+i > n){
				if(t[a-i]) cnt++;
				cout << cnt ;
			}
			printf(" %d\n",cnt);
		}
		 
	}
}

将错误的地方改正 

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

const int MAXN = 1000;
int t[MAXN];

int main()
{
	int n, a;
	while(scanf("%d%d",&n, &a)!=EOF){
		int cnt = 0;
		//memset(t, 0, sizeof(t));
		for(int i =1; i <= n; i++)
			cin >> t[i];
		if(t[a]) cnt++;
		for(int i = 1; i <= n; ++i){
			if(a-i > 0&&a+i <= n) {  //******改正错误点 
				if(t[a-i] == 1&&t[a+i] == 1)
					cnt+=2;	
			}
			else if(a-i <= 0&&a+i <= n){
				if(t[a+i])
					cnt++;
			}   
			else if(a-i > 0&&a+i > n){
				if(t[a-i])
					cnt++;	
			}			 
		}
		cout << cnt <<endl;
	}
}