codeforces555B

Case of Fugitive

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 adjacentislands. 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 + 1and 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 - 1numbers 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

Output

Yes
2 3 1

Input

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.

sol：把桥按照长度从小到大排序，对于每两座岛之间的桥长是一个闭区间[L,R]

对于桥一个个枚举过去，如当前桥长是a，对于所有L<=a的找到R最小的那个，贪心一遍即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=200005;
int n,m;
ll Daol[N],Daor[N];
struct Juli
{
    ll Down,Up,Id;
    inline bool operator<(const Juli &tmp)const
    {
        return Up>tmp.Up;
    }
}Dis[N];
inline bool cmp_Juli(Juli p,Juli q)
{
    return p.Down<q.Down;
}
priority_queue<Juli>heap;
struct Brg
{
    ll Len;
    int Id;
    inline bool operator<(const Brg &tmp)const
    {
        return Len<tmp.Len;
    }
}a[N];
int Ans[N];
int main()
{
    int i;
    R(n); R(m);
    for(i=1;i<=n;i++)
    {
        R(Daol[i]); R(Daor[i]);
        if(i>1) Dis[i-1]=(Juli){Daol[i]-Daor[i-1],Daor[i]-Daol[i-1],i-1};
    }
    sort(Dis+1,Dis+n,cmp_Juli);
    for(i=1;i<=m;i++) a[i]=(Brg){read(),i};
    sort(a+1,a+m+1);
    int Now=1,cnt=0;
//    for(i=1;i<n;i++) printf("%d %d\n",Dis[i].Down,Dis[i].Up);
//    puts("--------------------------");
    for(i=1;i<=m;i++)
    {
        while(Now<n&&a[i].Len>=Dis[Now].Down&&a[i].Len<=Dis[Now].Up) heap.push(Dis[Now++]);
        if(heap.empty()) continue;
        Juli tmp=heap.top(); heap.pop();
//        printf("%d %d %d\n",tmp.Down,tmp.Up,a[i].Len);
        if(tmp.Up<a[i].Len) return puts("No"),0;
        cnt++; Ans[tmp.Id]=a[i].Id;
    }
    if(cnt<n-1) return puts("No"),0;
    puts("Yes");
    for(i=1;i<n;i++) W(Ans[i]); puts("");
    return 0;
}
/*
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 

Input
6 9
1 4
10 18
23 29
33 43
46 57
59 77
11 32 32 19 20 17 32 24 32
Output
Yes
1 6 4 5 8 
*/

View Code

Case of Fugitive

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