CodeForces 555B Case of Fugitive：排序+贪心

CodeForces 555B：Case of Fugitive

Problem

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 $(2 ≤ n ≤ 2·10^5)$ and m $(1 ≤ m ≤ 2·10^5)$ — the number of islands and bridges.

Next n lines each contain two integers $l_i$ and $r_i$ $(1 ≤ l_i ≤ r_i ≤ 10^{18})$ — the coordinates of the island endpoints.

The last line contains m integer numbers $a_1, a_2, ..., a_m (1 ≤ a_i ≤ 1018)$ — 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 $b_1, b_2, ..., b_{n - 1}$, 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 #1

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

Output #1

Yes
2 3 1 

Input #2

2 2
11 14
17 18
2 9

Output #2

No

Input #3

2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999

Output #3

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.

Code

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=200005;
int n,m,cnt=0;
long long ln[maxn],rn[maxn];
struct IslandData{
long long L,R;
int id;
bool operator <(const IslandData b)const{
return R>b.R;
}
}isd[maxn];
struct BridgeData{
long long x;
int id;
bool operator <(const BridgeData b)const{
return x<b.x;
}
}a[maxn];
int id,x;
}ans[maxn];
priority_queue <IslandData> heap;
int ret=0,f=1;char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=getchar();
return ret*f;
}
long long ret=0,f=1;char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') f=(long long)-1;ch=getchar();}
while (ch>='0'&&ch<='9') ret=(long long)ret*(long long)10+(long long)(ch-'0'),ch=getchar();
return (long long)ret*f;
}
inline bool cmp(IslandData aa,IslandData bb){
return aa.L<bb.L;
}
return aa.id<bb.id;
}
int main(){
for (int i=1;i<n;i++){
IslandData now;
now.L=ln[i+1]-rn[i];
now.R=rn[i+1]-ln[i];
now.id=i;
isd[i]=now;
}
sort(a+1,a+1+m);
sort(isd+1,isd+n,cmp);
int j=1;
for (int i=1;i<=m;i++){
while (j<n&&isd[j].L<=a[i].x&&a[i].x<=isd[j].R) heap.push(isd[j++]);
if (heap.size()==0) continue;
IslandData now=heap.top();heap.pop();
if (now.R<a[i].x){
printf("No\n");
return 0;
}
ans[++cnt].id=now.id;ans[cnt].x=a[i].id;
}
if (cnt<n-1){printf("No\n");return 0;}
printf("Yes\n");
sort(ans+1,ans+1+cnt,cmp_id);
for (int i=1;i<=cnt;i++) printf("%d ",ans[i].x);
printf("\n");
return 0;
}

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