Codeforces Round #408 (Div. 2)

A. Buying A House

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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, ..., an that denote the availability and the prices of the houses. If house i is occupied, and therefore cannot be bought, then ai equals 0. Otherwise, house i can be bought, and ai 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, ..., an (0 ≤ ai ≤ 100) — denoting the availability and the prices of the houses.

It is guaranteed that am = 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

40

input

7 3 50
62 0 0 0 99 33 22

output

30

input

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

output

20

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 = 3and house 6 are only 30 meters away, while house m = 3 and house 7 are 40 meters away.

女神家在m，你要买一套房子里女神家最近，其中为0的房子是你买不起的，找你能买得起的离女神家房子最近的距离。

#include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int n,m,k;
cin>>n>>m>>k;
int mi=1<<30;
for(int i=1;i<=n;i++){
    int p;
    cin>>p;
    if(p&&i!=m&&p<=k)
        mi=min(mi,abs(i-m));
}
cout<<mi*10<<endl;
return 0;}

B. Find The Bone

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Zane the wizard is going to perform a magic show shuffling the cups.

There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i.

The problematic bone is initially at the position x = 1. Zane will confuse the audience by swapping the cups k times, the i-th time of which involves the cups at the positions x = ui and x = vi. If the bone happens to be at the position where there is a hole at any time, it will fall into the hole onto the ground and will not be affected by future swapping operations.

Do not forget that Zane is a wizard. When he swaps the cups, he does not move them ordinarily. Instead, he teleports the cups (along with the bone, if it is inside) to the intended positions. Therefore, for example, when he swaps the cup at x = 4 and the one at x = 6, they will not be at the position x = 5 at any moment during the operation.

Zane’s puppy, Inzane, is in trouble. Zane is away on his vacation, and Inzane cannot find his beloved bone, as it would be too exhausting to try opening all the cups. Inzane knows that the Codeforces community has successfully helped Zane, so he wants to see if it could help him solve his problem too. Help Inzane determine the final position of the bone.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 106, 1 ≤ m ≤ n, 1 ≤ k ≤ 3·105) — the number of cups, the number of holes on the table, and the number of swapping operations, respectively.

The second line contains m distinct integers h1, h2, ..., hm (1 ≤ hi ≤ n) — the positions along the x-axis where there is a hole on the table.

Each of the next k lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the positions of the cups to be swapped.

Output

Print one integer — the final position along the x-axis of the bone.

Examples

input

7 3 4
3 4 6
1 2
2 5
5 7
7 1

output

1

input

5 1 2
2
1 2
2 4

output

2

Note

In the first sample, after the operations, the bone becomes at x = 2, x = 5, x = 7, and x = 1, respectively.

In the second sample, after the first operation, the bone becomes at x = 2, and falls into the hole onto the ground.

B题看起来挺难的好像看不懂啊，其实就是简单的模拟，给你n个数，m个位置下有孔，然后再换，刚开始骨头在1号位置，有孔骨头就掉了。问最后骨头掉在哪里了

#include <bits/stdc++.h>
using namespace std;
const int N=1e6+10;
int a[N];
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int n,m,k;
cin>>n>>m>>k;
while(m--){
    int x;
    cin>>x;
    a[x]=1;
}
int ans=1;
while(k--){
    int x,y;
    cin>>x>>y;
    if(x==ans){
        if(!a[x])
        ans=y;
    }
    else if(y==ans){
        if(!a[y])
            ans=x;
    }
}
cout<<ans;
return 0;}