Problem Description

You've got a string a1,a2,…,an, consisting of zeros and ones.

Let's call a sequence of consecutive elements ai,ai?+?1,…,?aj (1≤?i≤?j≤?n) a substring of string a.

You can apply the following operations any number of times:

Choose some substring of string aa (for example, you can choose entire string) and reverse it, paying x coins for it (for example, ?0101101? →??0111001?);
Choose some substring of string aa (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying y coins for it (for example, ?0101101? → ?0110001?).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.

What is the minimum number of coins you need to spend to get a string consisting only of ones?

Input

The first line of input contains integers n, x and y?(1?≤?n?≤?300000,0≤x,y≤109) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).

The second line contains the string a of length n, consisting of zeros and ones.

Output

Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print 0, if you do not need to perform any operations.

Examples

Input

5 1 10
01000

Output

11

Input

5 10 1
01000

Output

2

Input

7 2 3
1111111

Output

0

題意：給出一個長度為 n 的二進制串，有兩種操作：將某一段反序，花費為 x；將某一段取反，花費為 y，求將這個串全都變為1的最小花費

思路：假設 0 的段數為cnt，則：如果 x<=y，那么就進行 cnt-1 次操作 1 和一次操作 2 ；如果 x>y，那么就進行 cnt 次操作 2?

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<set> #include<map> #include<stack> #include<ctime> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 100001 #define MOD 1e9+7 #define E 1e-6 #define LL long long using namespace std; int main() {int n,x,y;string str;cin>>n>>x>>y;cin>>str;LL cnt=0;for(LL i=0;i<n;i++)if(str[i]=='0'&&str[i+1]=='1')cnt++;if(str[n-1]=='0')cnt++;LL minn;if(x<y)minn=x*(cnt-1)+y;elseminn=y*cnt;if(cnt==0)minn=0;cout<<minn<<endl;return 0; }

?

總結

以上是生活随笔為你收集整理的Convert to Ones（CF-998C）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。