time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

【题解】



二分枚举需要的天数。

每次枚举的时候从最后一天往前处理。

第一次遇到的可以用来考试的天数。如果没考。则一定要用来考试。

然后再往前去找哪些天要用来复习。(即转换成给每个考试找复习时间);

如果最后有一些考试不能完成(没找到足够的复习天数);

则不行。

那么再把天数往右扩大；(l = m+1);

如果可行就减少天数(r = m-1);

#include <cstdio>

#include <queue>

using namespace std;

const int MAXN = 1e5 + 100;

int n, m;

int d[MAXN], a[MAXN],days = 0,temp[MAXN];

bool flag[MAXN];

queue <int> dl;

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i++)

        scanf("%d", &d[i]);

    for (int i = 1; i <= m; i++)

        scanf("%d", &a[i]),temp[i] = a[i];

    int l = 1, r = n;

    int best = -1;

    while (l <= r)

    {

        int tempm = m;

        for (int i = 1; i <= m; i++)

            flag[i] = false;

        for (int i = 1; i <= n; i++)

            a[i] = temp[i];

        while (!dl.empty())

            dl.pop();

        int m = (l + r) >> 1;

        for (int i = m; i >= 1; i--)

            if (!flag[d[i]] && d[i] != 0)

            {

                flag[d[i]] = true;

                dl.push(d[i]);

            }

            else

            {

                if (!dl.empty())

                {

                    int x = dl.front();

                    a[x]--;

                    if (!a[x])

                    {

                        dl.pop();

                        tempm--;

                    }

                }

            }

        if (tempm!=0)//不行->所有考试都完了才算通过。

        {

            l = m + 1;

        }

        else//行,尝试更优

        {

            r = m - 1;

            best = m;

        }

    }

    if (best == -1)

        puts("-1");

    else

        printf("%d\n", best);

    return 0;

}