Codeforces Round #408 (Div. 2) A.Buying A House【模拟】水题~

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在这个问题中，我们需要帮助一位名叫Zane的巫师找到离他心仪女孩最近且价格合适的房子。通过给定的房子价格列表和Zane的预算，我们要找出最短的距离。

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Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.

$\text{[math]}$

The girl lives in house m of a village. There are n houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house n. The village is also well-structured: house i and house i + 1 (1 ≤ i < n) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.

You will be given n integers a₁, a₂, ..., a_n that denote the availability and the prices of the houses. If house i is occupied, and therefore cannot be bought, then a_i equals 0. Otherwise, house i can be bought, and a_i represents the money required to buy it, in dollars.

As Zane has only k dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 100, 1 ≤ m ≤ n, 1 ≤ k ≤ 100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.

The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 100) — denoting the availability and the prices of the houses.

It is guaranteed that a_m = 0 and that it is possible to purchase some house with no more than k dollars.

Output

Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.

Examples

Input

5 1 20
0 27 32 21 19

Output

Input

7 3 50
62 0 0 0 99 33 22

Output

Input

10 5 100
1 0 1 0 0 0 0 0 1 1

Output

Note

In the first sample, with k = 20 dollars, Zane can buy only house 5. The distance from house m = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.

In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house m = 3 and house 6 are only 30 meters away, while house m = 3 and house 7 are 40 meters away.

题目大意：

给你N个房子，每个房子的价格为a【i】.a【i】==0的表示不能购买，现在他的位子在m，有k钱.他想买一个房子，使得这个房子尽可能的靠近m这个位子。

现在设定两个房子间的距离为10.

思路：

直接暴力模拟即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int a[150];
int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        int output=0x3f3f3f3f;
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=m;i<=n;i++)
        {
            if(a[i]!=0)
            {
                if(a[i]<=k)output=min(output,(i-m)*10);
            }
        }
        for(int i=m;i>=1;i--)
        {
            if(a[i]!=0)
            {
                if(a[i]<=k)output=min(output,(m-i)*10);
            }
        }
        printf("%d\n",output);
    }
}