codeforce 732e Sockets （贪心）

The ICM ACPC World Finals is coming! Unfortunately, the organizers of the competition were so busy preparing tasks that totally missed an important technical point — the organization of electricity supplement for all the participants workstations.

There are n computers for participants, thei-th of which has power equal to positive integerp_i. At the same time there arem sockets available, the j-th of which has power euqal to positive integer s_j. It is possible to connect thei-th computer to the j-th socket if and only if their powers are the same: p_i = s_j. It is allowed to connect no more than one computer to one socket. Thus, if the powers of all computers and sockets are distinct, then no computer can be connected to any of the sockets.

In order to fix the situation professor Puch Williams urgently ordered a wagon of adapters — power splitters. Each adapter has one plug and one socket with a voltage divider between them. After plugging an adapter to a socket with powerx, the power on the adapter's socket becomes equal to, it means that it is equal to the socket's power divided by two with rounding up, for example and.

Each adapter can be used only once. It is possible to connect several adapters in a chain plugging the first to a socket. For example, if two adapters are plugged one after enother to a socket with power10, it becomes possible to connect one computer with power3 to this socket.

The organizers should install adapters so that it will be possible to supply with electricity the maximum number of computersc at the same time. If there are several possible connection configurations, they want to find the one that uses the minimum number of adaptersu to connect c computers.

Help organizers calculate the maximum number of connected computers c and the minimum number of adapters u needed for this.

The wagon of adapters contains enough of them to do the task. It is guaranteed that it's possible to connect at least one computer.

Input

The first line contains two integers n andm (1 ≤ n, m ≤ 200 000) — the number of computers and the number of sockets.

The second line contains n integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ 10⁹) — the powers of the computers.

The third line contains m integers s₁, s₂, ..., s_m (1 ≤ s_i ≤ 10⁹) — the power of the sockets.

Output

In the first line print two numbers c andu — the maximum number of computers which can at the same time be connected to electricity and the minimum number of adapters needed to connectc computers.

In the second line print m integers a₁, a₂, ..., a_m (0 ≤ a_i ≤ 10⁹), where a_i equals the number of adapters orginizers need to plug into thei-th socket. The sum of all a_i should be equal to u.

In third line print n integers b₁, b₂, ..., b_n (0 ≤ b_i ≤ m), where the b_j-th equals the number of the socket which thej-th computer should be connected to. b_j = 0 means that the j-th computer should not be connected to any socket. All b_j that are different from0 should be distinct. The power of the j-th computer should be equal to the power of the socket b_j after plugging in a_bj adapters. The number of non-zerob_j should be equal toc.

If there are multiple answers, print any of them.

Examples

Input

2 21 12 2

Output

2 21 11 2

Input

2 12 10099

Output

1 661 0

题意：有n个电脑和m个适配器(n,m<=2e5) 他们各自都有自己的电压，要求电脑和适配器电压相同才匹配。对适配器的一次操作可以将其电压除以2向上取整。求最多有多少个电脑可以匹配到适配器，在这个基础上，再求最少操作总次数，并输出方案。

思路：贪心策略，把每台电脑从小到大排，电压从小到大排，然后暴力枚举变压器的个数，最大用30个，所以不会增加时间的复杂度。从0~30一次枚举过，然后在O（n+m）枚举过去匹配电脑跟电压。。。。。脑洞题。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 2e5+100;int a[N],b[N],vis_p[N],vis_s[N];struct node{int x,index;}p[N],s[N];bool cmp(node t1,node t2){if(t1.x<t2.x) return true;return false;}int main(){int n,m,c,u,i,j,num;scanf("%d%d",&n,&m);for(i=1;i<=n;i++) {scanf("%d",&p[i].x);p[i].index=i;}for(i=1;i<=m;i++) {scanf("%d",&s[i].x);s[i].index=i;}sort(p+1,p+1+n,cmp);sort(s+1,s+1+m,cmp);int p1,s1;c=u=0;for(i=0;i<=30;i++) {p1=s1=1;while(p1<=n&&s1<=m) {if(vis_p[p1]) p1++;else if(vis_s[s1]) s1++;else if(s[s1].x==p[p1].x) {c++;     //电脑可用数 u+=i;    //变压器数 vis_p[p1]=1;//该电脑匹配完成 vis_s[s1]=1;//a[s1]+=i; //s1号电压要用的变压器个数 a[s[s1].index]+=i;//b[p1]=s1; b[p[p1].index]=s[s1].index;p1++;s1++;}else if(s[s1].x>p[p1].x) p1++;else s1++;}if(s[m].x==1) break;for(j=1;j<=m;j++) s[j].x=s[j].x-(s[j].x>>1); }printf("%d %d\n",c,u);for(i=1;i<=m;i++) printf("%d ",a[i]);printf("\n");for(i=1;i<=n;i++) printf("%d ",b[i]);printf("\n");return 0;}