Codeforces 814A-An abandoned sentiment from past

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 k equals 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 bshould 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 brespectively.

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

4 211 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 are 5 and 4.
• There are two possible resulting sequences: 11, 5, 4, 14 and 11, 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.

题意：给你一个序列，有n个数，里面有m个0，给你m个非零数，每个数只能用一次，向序列里填数（这些数字全不相同），问能否使得序列是非递增的

解题思路：当m大于1时，一定可以使得序列非递增，当m为0时，只要判断原序列是否非递增，当m为1时，将数字填进去后判断序列是否非递增

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m;int a[300], b[300];int main(){while (~scanf("%d%d", &n, &m)){a[0] = -1;for (int i = 1; i <= n; i++) scanf("%d", &a[i]);a[n + 1] = INF;for (int i = 1; i <= m; i++) scanf("%d", &b[i]);int k = -1,flag=0;for (int i = 1; i <= n; i++){if (!a[i]) continue;if (a[i] < k) flag = 1;k = a[i];}if (!flag&&m==0) printf("No\n");else if (!flag&&m == 1){int flag1 = 0;for (int i = 1; i <= n; i++)if (!a[i] && b[1] > a[i - 1]&&b[1]<a[i+1]) flag1 = 1;if (flag1) printf("No\n");else printf("Yes\n");}else printf("Yes\n");}return 0;}