Problem: One day Natalia was walking in the woods when she met a little mushroom gnome. The gnome told her the following story:

Everybody knows that the mushroom gnomes' power lies in the magic mushrooms that grow in the native woods of the gnomes. There are n trees and m magic mushrooms in the woods: the i-th tree grows at a point on a straight line with coordinates ai and has the height of hi, the j-th mushroom grows at the point with coordinates bj and has magical powers zj.

But one day wild mushroommunchers, the sworn enemies of mushroom gnomes unleashed a terrible storm on their home forest. As a result, some of the trees began to fall and crush the magic mushrooms. The supreme oracle of mushroom gnomes calculated in advance the probability for each tree that it will fall to the left, to the right or will stand on. If the tree with the coordinate x and height h falls to the left, then all the mushrooms that belong to the right-open interval [x - h, x), are destroyed. If a tree falls to the right, then the mushrooms that belong to the left-open interval (x, x + h] are destroyed. Only those mushrooms that are not hit by a single tree survive.

Knowing that all the trees fall independently of each other (i.e., all the events are mutually independent, and besides, the trees do not interfere with other trees falling in an arbitrary direction), the supreme oracle was also able to quickly calculate what would be the expectation of the total power of the mushrooms which survived after the storm. His calculations ultimately saved the mushroom gnomes from imminent death.

Natalia, as a good Olympiad programmer, got interested in this story, and she decided to come up with a way to quickly calculate the expectation of the sum of the surviving mushrooms' power.

Input

The first line contains two integers n and m (1 ≤ n ≤ 105, 1 ≤ m ≤ 104) — the number of trees and mushrooms, respectively.

Each of the next n lines contain four integers — ai, hi, li, ri (|ai| ≤ 109, 1 ≤ hi ≤ 109, 0 ≤ li, ri, li + ri ≤ 100) which represent the coordinate of the i-th tree, its height, the percentage of the probabilities that the tree falls to the left and to the right, respectively (the remaining percentage is the probability that the tree will stand on).

Each of next m lines contain two integers bj, zj (|bj| ≤ 109, 1 ≤ zj ≤ 103) which represent the coordinate and the magical power of the j-th mushroom, respectively.

An arbitrary number of trees and mushrooms can grow in one point.

Output

Print a real number — the expectation of the total magical power of the surviving mushrooms. The result is accepted with relative or absolute accuracy 10 - 4.

Examples

Input

1 1
2 2 50 50
1 1


Output

0.5000000000


Input

2 1
2 2 50 50
4 2 50 50
3 1


Output

0.2500000000

Note

It is believed that the mushroom with the coordinate x belongs to the right-open interval [l, r) if and only if l ≤ x < r. Similarly, the mushroom with the coordinate x belongs to the left-open interval (l, r] if and only if l < x ≤ r.

In the first test the mushroom survives with the probability of 50%, depending on where the single tree falls.

In the second test the mushroom survives only if neither of the two trees falls on it. It occurs with the probability of 50%  ×  50% = 25%.

Pretest №12 is the large test with 105 trees and one mushroom.

Solution (in JAVA):

Solution:
import java.io.* ;
import java.util.*;

import static java.lang.Math.* ;
import static java.util.Arrays.* ;

public class C {
	
	public static void main(String[] args) throws IOException {
		
		new C().solveProblem();
		
		out.close();
	}

	static Scanner in = new Scanner(new InputStreamReader(System.in));
	//static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	static PrintStream out = new PrintStream(new BufferedOutputStream(System.out));
	
	int n, m ;
	
	public void solveProblem() {		

		n = in.nextInt() ;
		m = in.nextInt() ;
		
		Interval[] left = new Interval[n] ;
		Interval[] right = new Interval[n] ;
		
		for( int i = 0 ; i < n ; i++ ){
			int a = in.nextInt() ;
			int h =in.nextInt() ;
			int l = 100-in.nextInt() ;
			int r = 100-in.nextInt() ;
			
			left[i] = new Interval(a,h,l) ;
			right[i] = new Interval(a,h,r) ;
		}
		
		Mushroom[] rooms = new Mushroom[m] ;
		double power = 0.0 ;
		for( int i = 0 ; i < m ; i++ ){
			rooms[i] = new Mushroom(in.nextInt(),in.nextInt() ) ;
			power += rooms[i].p ;
		}
		sort( left ) ;
		sort( right ) ;
		sort( rooms ) ;
		
		double[] dood = new double[m] ;
		
		double kans = 1.0 ; // kans dat hij leeft
		
		PriorityQueue<Remove> qu = new PriorityQueue<Remove>() ;
		int j = 0 ;
		for( int i = 0 ; i < m ; i++ ){
			
			while( j < n && right[j].x < rooms[i].x ){
				if(right[j].p == 0 ){
					j++ ;
					continue ;					
				}
				
				if( right[j].x+right[j].h >= rooms[i].x ){
				kans = kans*((double) right[j].p/100.0) ; // 1 - de kans dat hij overleeft
				qu.add(new Remove(right[j].x+right[j].h, right[j].p)) ;
				}
				j++ ;
			}
			
			while( !qu.isEmpty() && qu.peek().x < rooms[i].x){
				kans = kans / (qu.peek().p/100.0) ;
				qu.poll();
			}

	
			dood[i] = (1.0-kans) ;
			//power -= (1.0-kans)*rooms[i].p ;
			
		}
		
		kans = 1.0 ;
		
		PriorityQueue<Remove2> quu = new PriorityQueue<Remove2>() ;
		j = n-1 ;
		for( int i = m-1 ; i >=0 ; i-- ){
			
			while( j >= 0 && left[j].x > rooms[i].x ){
				
				if(left[j].p == 0 ){
					j-- ;
					continue ;
				}
				if( left[j].x-left[j].h <= rooms[i].x){
					kans = kans*((double) left[j].p/100.0) ; // 1 - de kans dat hij overleeft
					quu.add(new Remove2(left[j].x-left[j].h, left[j].p)) ;
				}
				j-- ;
			}
			
			while( !quu.isEmpty() && quu.peek().x > rooms[i].x){
				kans = kans / (quu.peek().p/100.0) ;
				quu.poll();
			}
			
			
			dood[i] = 1 - (1.0-dood[i] )*kans ;
			//power -= (1.0-kans)*rooms[i].p ;
			
		}
		
		
		qu = new PriorityQueue<Remove>() ;
		j = 0 ;
		for( int i = 0 ; i < m ; i++ ){
			
			while( j < n && right[j].x < rooms[i].x ){
				if(right[j].p == 0 )
					qu.add(new Remove(right[j].x+right[j].h, right[j].p)) ;
				j++ ;
			}
			
			while( !qu.isEmpty() && qu.peek().x < rooms[i].x){
				qu.poll();
			}

			
			if( qu.size() > 0 )
				dood[i] =1.0 ;
			//power -= (1.0-kans)*rooms[i].p ;
			
		}
		
		
		quu = new PriorityQueue<Remove2>() ;
		j = n-1 ;
		for( int i = m-1 ; i >=0 ; i-- ){
			
			while( j >= 0 && left[j].x > rooms[i].x ){
				
				if(left[j].p == 0 )					
					quu.add(new Remove2(left[j].x-left[j].h, left[j].p)) ;
				j-- ;
			}
			
			while( !quu.isEmpty() && quu.peek().x > rooms[i].x){
				quu.poll();
			}
			
			
			
			if( quu.size() > 0 )
				dood[i] =1.0 ;
			//power -= (1.0-kans)*rooms[i].p ;
			
		}
		
		for( int i = 0 ; i < m ; i++ )
			power -= dood[i]*(double) rooms[i].p ;
		out.println(power) ;
	}

	static int[] readGet( int n ){
	  	int[] get = new int[n] ;
	  	for( int i = 0 ; i < n ; i++ )
	  		get[i] = in.nextInt();
	  	return get ;
	}
	
	class Remove implements Comparable<Remove>{
		int x ;
		double p ;
		
		public Remove( int x, int p){
			this.x = x ;
			this.p = p ;
		}
		
		public int compareTo( Remove o){
			return x - o.x ;
		}
	}
	
	class Remove2 implements Comparable<Remove2>{
		int x ;
		double p ;
		
		public Remove2( int x, int p){
			this.x = x ;
			this.p = p ;
		}
		
		public int compareTo( Remove2 o){
			return o.x-x  ;
		}
	}
	class Interval implements Comparable<Interval>{
		
		int x ;
		int h ;
		
		int p ;
		
		public Interval( int x, int y, int p ){
			this.x = x ;
			this.h = y ;
			this.p = p ;
		}
		
		public int compareTo( Interval o ){
			return x - o.x ;
		}
	}
	class Mushroom implements Comparable<Mushroom>{
		int x ;
		int p ;
		
		public Mushroom( int x, int p ){
			this.x = x ;
			this.p = p ;
		}
		
		public int compareTo( Mushroom o ){
			return x - o.x ;
		}
	}
	
       
}