Today CodeFamer is going to cut trees.There are N trees standing in a line. They are numbered from 1 to N. The tree numbered i has height hi. We say that two uncutted trees whose numbers are x and y are in the same block if and only if they are fitting in one of blow rules:

1)x+1=y or y+1=x;

2)there exists an uncutted tree which is numbered z, and x is in the same block with z, while y is also in the same block with z.

Now CodeFamer want to cut some trees whose height is not larger than some value, after those trees are cut, how many tree blocks are there?

Multi test cases (about 15).

For each case, first line contains two integers N and Q separated by exactly one space, N indicates there are N trees, Q indicates there are Q queries.

In the following N lines, there will appear h[1],h[2],h[3],…,h[N] which indicates the height of the trees.

In the following Q lines, there will appear q[1],q[2],q[3],…,q[Q] which indicates CodeFamer’s queries.

Please process to the end of file.

[Technical Specification]

1≤N,Q≤50000

0≤h[i]≤1000000000(109)

0≤q[i]≤1000000000(109)

For each q[i], output the number of tree block after CodeFamer cut the trees whose height are not larger than q[i].

3 2

5

2

3

6

2

0

2

 #include<stdio.h>

 #include<map>

 #include<algorithm>

 #include<set>

 #include<string.h>

 using namespace std;

 int n , m ;

 const int M =  +  ;

 int cut[M] ;

 bool vis[M] ;

 pair <int , int> a[M] ;

 void solve ()

 {

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

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

         a[i].second = i ;

     }

     sort (a +  , a + n + ) ;

     for (int i = n ; i >=  ; i --) {

         cut [i] = cut [i + ] ;

         int now = a[i].second ;

         if (vis[now - ] && vis[now + ] ) {

             cut[i] -- ;

         }

         else if (!vis[now - ] && !vis[now + ]) {

             cut[i] ++ ;

         }

         vis[now] =  ;//1 代表没砍 ， 0 代表砍了

     }

     while (m--) {

         int h ;

         scanf ("%d" , &h) ;

         int x = upper_bound (a +  , a + n +  , make_pair (h , n + ) ) - a ;

         if (x > n) puts ("") ; else printf ("%d\n" , cut[x]) ;

     }

 }

 int main ()

 {

    // freopen ("a.txt" , "r" , stdin ) ;

     while (~ scanf ("%d%d" , &n , &m)) {

         memset (vis ,  , sizeof(vis)) ;

         memset (cut ,  , sizeof(cut)) ;

         solve () ;

     }

     return  ;

 }