D. Exams

time limit per test

memory limit per test

input

output

n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.

m

On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.

ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.

Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.

Input

n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.

n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.

m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.

Output

-1.

Examples

input

7 2
0 1 0 2 1 0 2
2 1

output

5

input

10 3
0 0 1 2 3 0 2 0 1 2
1 1 4

output

9

input

5 1
1 1 1 1 1
5

output

-1


贪心。


#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
const int maxn=1e5;
int n,m;
int a[maxn+5];
int d[maxn+5];
int vis[maxn+5];
int check(int x)
{
  int num=0;
  int need=0;
  memset(vis,0,sizeof(vis));
  for(int i=x;i>=1;i--)
  {
    if(!vis[a[i]]&&num!=m&&a[i]!=0)
    {
      need+=d[a[i]];
      num++;
      vis[a[i]]=1;
      continue;
    }
    if(need)
    {
      need--;
    }
  }
  if(need<=0&&num==m) return 1;
  return 0;
}
int main()
{
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    scanf("%d",&a[i]);
  for(int i=1;i<=m;i++)
    scanf("%d",&d[i]);
  int l=1,r=n;
  int ans=-1;
  while(l<=r)
  {
    int mid=(l+r)>>1;
    if(check(mid))
    {
      ans=mid;
      r=mid-1;
    }
    else
      l=mid+1;
  }
  
    printf("%d\n",ans);
  return 0;
}