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 [li, ri], besides, ri < li + 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 li ≤ x ≤ ri, li + 1 ≤ y ≤ ri + 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.

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef unsigned long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
inline ll read()
{int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}//**************************************************************************************set<pair<ll,int> >s;
pair<pair<ll,ll>,int> br[maxn];
pair<ll,ll> a[maxn];
set<pair<ll,int> >::iterator it;
int ans[maxn];
int main()
{int n=read(),m=read();for(int i=0;i<n;i++)cin>>a[i].first>>a[i].second;for(int i=0;i<n-1;i++)br[i]=make_pair(make_pair(a[i].second-a[i+1].first,a[i].first-a[i+1].second),i);for(int i=0;i<m;i++){ll kiss;cin>>kiss;s.insert(make_pair(kiss,i));}sort(br,br+n-1);for(int i=0;i<n-1;i++){ll l=-br[i].first.first,r=-br[i].first.second;it=s.lower_bound(make_pair(r,inf));if(it==s.begin())return puts("No");it--;if(it->first<l)return puts("No");ans[br[i].second]=it->second;s.erase(it);}puts("Yes");for(int i=0;i<n-1;i++)cout<<ans[i]+1<<" ";
}