codeforces 814 A 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 sequence b, whose length kequals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b 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 from b is used exactly once, and the resulting sequence is not 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 sequence a and b respectively.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.

The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.

Input guarantees that apart from 0, no integer occurs in a and b more than once in total.

Output

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

Examples

input

4 2
11 0 0 14
5 4

output

Yes

input

6 1
2 3 0 8 9 10
5

output

No

input

4 1
8 94 0 4
89

output

Yes

input

7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7

output

Yes

输入n，m，

n表示数字序列长度，序列中0表示该位数字丢失m表示丢失数字个数

第一行n个数字表示该数字序列，下一行表示m个丢失数字，将0用丢失数字替换（不必按顺序替换），如果能构成非递增序列则输出yes否则no

同时题目给出没有任何两个相同的数（除0以外）

因此如果丢失两个以上的数则必定可以构成非递增序列

#include <iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int a[110],b[110];
    int n,m;
    scanf("%d%d",&n,&m);
   for(int i=0;i<n;i++)
   {
       scanf("%d",&a[i]);
   }
   for(int i=0;i<m;i++)
   {
       scanf("%d",&b[i]);
   }
   int h=0;
   if(m>=2)
    printf("Yes\n"),h=1;
   else
   {
       for(int i=0;i<n;i++)
       {
           if(a[i]==0)
            a[i]=b[0];
            if(a[i]<a[i-1])//判断是否为递增序列
            {
                printf("Yes\n");
                h=1;
                break;
            }
       }
      if(h==0)
      printf("No\n");
   }
    return 0;
}