An abandoned sentiment from past

A. An abandoned sentiment from past

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.

To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.

Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequenceb, whose lengthk equals the number of lost elements ina (i.e. the number of zeros). Hitagi is to replace each zero ina with an element fromb so that each element in b should be used exactly once. Hitagi knows, however, that,apart from0, no integer occurs ina andb more than once in total.

If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer fromb is used exactly once, and the resulting sequence isnot increasing.

Input

The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequencea andb respectively.

The second line contains n space-separated integersa₁, a₂, ..., a_n (0 ≤ a_i ≤ 200) — Hitagi's broken sequence with exactlyk zero elements.

The third line contains k space-separated integersb₁, b₂, ..., b_k (1 ≤ b_i ≤ 200) — the elements to fill into Hitagi's sequence.

Input guarantees that apart from 0, no integer occurs ina andb more than once in total.

Output

Output "Yes" if it's possible to replace zeros ina with elements inb and make the resulting sequence not increasing, and "No" otherwise.

Examples

Input

4 211 0 0 145 4

Output

Yes

Input

6 12 3 0 8 9 105

Output

No

Input

4 18 94 0 489

Output

Yes

Input

7 70 0 0 0 0 0 01 2 3 4 5 6 7

Output

Yes

Note

In the first sample:

• Sequence a is 11, 0, 0, 14.
• Two of the elements are lost, and the candidates in b are5 and4.
• There are two possible resulting sequences: 11, 5, 4, 14 and11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".

In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.

思路:当k>1时，由于数字不一样，一定会有一种方法使得数列非递增。当k=0或k=1,只需判断数列是否是递增的即可。

代码:

#include <iostream>#include <cstdio>using namespace std;int a[100];int main(){    int n,k,b;    while(cin>>n>>k)    {       int index;       for(int i=0;i<n;i++)       {           cin>>a[i];           if(a[i]==0) index=i;       }       for(int i=0;i<k;i++)       {           cin>>b;           a[index]=b;       }       if(k>1){          cout<<"Yes"<<endl;          continue;       }       else{            int flag=0;           for(int i=0;i<n-1;i++)           {               if(a[i]>a[i+1]){                   cout<<"Yes"<<endl;                   flag=1;                   break;               }           }         if(!flag) cout<<"No"<<endl;       }    }    return 0;}