Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
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.
About each subject Vasiliy know a number 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
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains 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.
The third line contains 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
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -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
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
#include<bits/stdc++.h>
using namespace std;
#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define read(x,y) scanf("%d%d",&x,&y)
#define root l,r,rt
#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn =1e5+5;
/*
题目大意:给定一系列课程，
每个课程都有复习天数，和考试的预定时间（多个），
问最少多少天所有考试全部考完。
天哪，想了好久，二分想的到，
维护一个最小值用l端即可，
但判断中的贪心就难想了/。。。
倒着推，遇到空白天，则压力值减一，（如果有的话）
如果遇到之前考完的试则直接压力减一，
如果遇到考试则累加压力值，如果后面的天数不足以支撑现在的考试复习则直接返回false。
*/
int seq[maxn];
int sub[maxn];
int n,m;
int vis[maxn];///是否已经遍历过
bool judge(int x)
{
int left=0;
memset(vis,0,sizeof(vis));
for(int i=x;i>=1;i--)
{
if(vis[seq[i]] || !seq[i])///遇到空白天就减去压力值，如果有的话
{
if(left) left--;
continue;
}
vis[seq[i]]=1;///遇到考试就考
if(i-left-1<sub[seq[i]]) return false;///天数不够用
left+=sub[seq[i]];
}
for(int i=1;i<=m;i++) if(!vis[i]) return false;
return true;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) scanf("%d",&seq[i]);
for(int i=1;i<=m;i++) scanf("%d",&sub[i]);
int l=1,r=n+1;
while(l<r)
{
int mid=l+r>>1;
if(judge(mid)) r=mid;
else l=mid+1;
}
if(l>n) puts("-1");
else cout<<l<<endl;
///cout<<ans<<endl;
return 0;
}