CodeForces - 556D

D. Case of Fugitive

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.

The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [l_i, r_i], besides, r_i < l_i + 1 for 1 ≤ i ≤ n - 1.

To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that l_i ≤ x ≤ r_i, l_i + 1 ≤ y ≤ r_i + 1 and y - x = a.

The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.

Input

The first line contains integers n (2 ≤ n ≤ 2·10⁵) and m (1 ≤ m ≤ 2·10⁵) — the number of islands and bridges.

Next n lines each contain two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10¹⁸) — the coordinates of the island endpoints.

The last line contains m integer numbers a₁, a₂, ..., a_m (1 ≤ a_i ≤ 10¹⁸) — the lengths of the bridges that Andrewid got.

Output

If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b₁, b₂, ..., b_n - 1, which mean that between islands i and i + 1 there must be used a bridge number b_i.

If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.

Examples

Input

4 4
1 4
7 8
9 10
12 14
4 5 3 8

Output

Yes
2 3 1 

Input

2 2
11 14
17 18
2 9

Output

No

Input

2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999

Output

Yes
1 

Note

In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.

In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.

题意：n-1个区间 m个数

数要满足的条件是大于等于左区间 小于等于又区间  问着m个数中有没有n-1个数满足这些区间（一个数不能重复满足多个区间）

区间按左从小到大排序 用个优先队列维护区间右边的值

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<string>
 #include<queue>
 #include<algorithm>
 #include<stack>
 #include<cstring>
 #include<vector>
 #include<list>
 #include<bitset>
 #include<set>
 #include<map>
 #include<time.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const int N=+;
 const int inf=0x3f3f3f3f;
 struct node{
     ll l,r;
     int pos;
     friend bool operator<(node aa,node bb){
         return aa.r>bb.r;
     }
 }a[N];
 struct Node{
     ll x;int pos;
 }b[N];
 bool cmp1(node aa,node bb){
     return aa.l<bb.l;
 }
 bool cmp2(Node aa,Node bb){
     return aa.x<bb.x;
 }
 priority_queue<node>q;
 int ans[N];
 int main(){
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF){
         ll l,r;
         scanf("%I64d%I64d",&l,&r);
         int t=;
         ll x,y;
         for(int i=;i<n;i++){
             scanf("%I64d%I64d",&x,&y);
             ll t1=x;
             ll t2=y;
             a[++t].l=x-r;
             a[t].r=y-l;
             a[t].pos=i;
             l=t1;r=t2;
         }
         sort(a+,a+t+,cmp1);
         for(int i=;i<=m;i++){
             scanf("%I64d",&b[i].x);
             b[i].pos=i;
         }
         sort(b+,b++m,cmp2);
         int k=;
         int flag=;
         int cnt=;
         for(int i=;i<=m;i++){
             x=b[i].x;
             for(int j=k;j<=t;j=k){
                 if(a[j].l<=x&&a[j].r>=x){
                     q.push(a[j]);k++;
                 }
                 else{
                     break;
                 }
             }
             if(q.empty())continue;
             node aa=q.top();q.pop();
             if(x>aa.r){
                 flag=;
                 break;
             }
             cnt++;
             //cout<<cnt<<endl;
             ans[aa.pos]=b[i].pos;
         }
         if(cnt<t||flag){
             cout<<"No"<<endl;continue;
         }
         cout<<"Yes"<<endl;
         for(int i=;i<=t;i++)cout<<ans[i]<<" ";
         cout<<endl;
     }
 }