One day Vasya went out for a walk in the yard but there weren't any of his friends outside and he had no one to play touch and run. But the boy didn't lose the high spirits and decided to play touch and run with himself. You may ask: "How did he do that?" The answer is simple.

Vasya noticed that the yard is a rectangular n × m field. The squares have coordinates (x, y) (1 ≤ x ≤ n, 1 ≤ y ≤ m), where x is the index of the row and y is the index of the column.

Initially Vasya stands in the square with coordinates (xc, yc). To play, he has got a list of k vectors (dxi, dyi) of non-zero length. The game goes like this. The boy considers all vectors in the order from 1 to k, and consecutively chooses each vector as the current one. After the boy has chosen a current vector, he makes the maximally possible number of valid steps in the vector's direction (it is possible that he makes zero steps).

A step is defined as one movement from the square where the boy is standing now, in the direction of the current vector. That is, if Vasya is positioned in square (x, y), and the current vector is (dx, dy), one step moves Vasya to square (x + dx, y + dy). A step is considered valid, if the boy does not go out of the yard if he performs the step.

Vasya stepped on and on, on and on until he ran out of vectors in his list. Ha had been stepping for so long that he completely forgot how many steps he had made. Help the boy and count how many steps he had made.

Input

The first input line contains two integers n and m (1 ≤ n, m ≤ 109) — the yard's sizes. The second line contains integers xc and yc — the initial square's coordinates (1 ≤ xc ≤ n, 1 ≤ yc ≤ m).

The third line contains an integer k (1 ≤ k ≤ 104) — the number of vectors. Then follow k lines, each of them contains two integers dxi and dyi (|dxi|, |dyi| ≤ 109, |dx| + |dy| ≥ 1).

Output

Print the single number — the number of steps Vasya had made.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

4 5
1 1
3
1 1
1 1
0 -2


Output

4


Input

10 10
1 2
1
-1 0


Output

0

Note

In the first sample Vasya is initially positioned at square (1, 1) and makes 3 steps by the first vector (1, 1). So, he consecutively visits the squares (2, 2), (3, 3), (4, 4). Then he makes 0 steps by the second vector (1, 1). He makes 1 more step by the third vector (0, - 2) and he ends up in square (4, 2). Overall, Vasya makes 4 steps.

In the second sample Vasya is initially positioned in square (1, 2) and makes 0 steps by vector ( - 1, 0), as the square with coordinates (0, 2) is located outside the yard.

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */





//import com.sun.xml.internal.ws.util.StringUtils;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
import java.util.Vector;
import java.util.stream.Collectors;





/**
 *
 * @author george
 */

public class main {
    /*
public static boolean isPrime(long n) {
    if(n < 2) return false;
    if(n == 2 || n == 3) return true;
    if(n%2 == 0 || n%3 == 0) return false;
    long sqrtN = (long)Math.sqrt(n)+1;
    for(long i = 6L; i <= sqrtN; i += 6) {
        if(n%(i-1) == 0 || n%(i+1) == 0) return false;
    }
    return true;
}*/
/*
    private static long f(long l) {
        throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
    }

public static boolean contains(final int[] arr, final int key) {
    return Arrays.stream(arr).anyMatch(i -> i == key);
}
 static boolean isSubSequence(String str1, String str2, int m, int n)
    {
       
        if (m == 0) 
            return true;
        if (n == 0) 
            return false;
             
        
        if (str1.charAt(m-1) == str2.charAt(n-1))
            return isSubSequence(str1, str2, m-1, n-1);
 
        
        return isSubSequence(str1, str2, m, n-1);
    }

    static int gcdThing(int a, int b) {
    BigInteger b1 = BigInteger.valueOf(a);
    BigInteger b2 = BigInteger.valueOf(b);
    BigInteger gcd = b1.gcd(b2);
    return gcd.intValue();
}
    public static boolean checkAnagram(String str1, String str2) {
        int i=0;
        for (char c : str1.toCharArray()) {
        i = str2.indexOf(c, i) + 1;
        if (i <= 0) { return false; }
    }
    return true;

}
    
    public static Set primeFactors(int num)
    {
        Set<Integer>Prime=new HashSet<Integer>();
        // Print the number of 2s that divide n
        int tw=2;
        while (num%2==0)
        {
            Prime.add(tw);
            num /= 2;
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(num); i+= 2)
        {
            // While i divides n, print i and divide n
            while (num%i == 0)
            {
                Prime.add(i);
                num/= i;
            }
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (num> 2)
            Prime.add(num);        
        return Prime;
    }
    public static String Winner(int [] arr)
    {
        int maxsum=-1;
        int temp=0;
        List<Integer> L=Arrays.stream(arr).boxed().collect(Collectors.toList());
        int count=0;
        String winner="First";
        while(L.isEmpty()!=true){
           if(count%2==0){winner="First";
           //take max odd interval
            for (int i = 0; i < L.size(); i++) {
                
            }
            }
           else{winner="Second";
           
           
           
           
           
           
           
           }
           count++;
        }
    
    
    return winner;
    }
    public static String changeCharInPosition(int position, char ch, String str){
    char[] charArray = str.toCharArray();
    charArray[position] = ch;
    return new String(charArray);
    }
    public static boolean CheckCanObtainYfromX(String y,String x){
        boolean cond=true;
        String help="";
        for (int i = 0; i < y.length(); i++) {
            help="";help+=y.charAt(i);
            if(y.contains(help)==false){x=x.replaceFirst(help,"A");cond=false;break;}
            
        }
        return cond;
    }
   public static int LongestPalindrome(String x){
    int n=x.length();
    boolean table[][] = new boolean[n][n];
    int maxLength = 1;
     for (int i = 0; i < n; ++i)
            table[i][i] = true;
   int start = 0;
    for (int i = 0; i < n - 1; ++i) {
            if (x.charAt(i) == x.charAt(i + 1)) {
                table[i][i + 1] = true;
                start = i;
                maxLength = 2;
            }
        }
     for (int k = 3; k <= n; ++k) {
            for (int i = 0; i < n - k + 1; ++i) 
            {
                int j = i + k - 1;
               if (table[i + 1][j - 1] && x.charAt(i) == x.charAt(j)) {
                    table[i][j] = true;
                if (k > maxLength) {
                        start = i;
                        maxLength = k;
                    }
                }
            }
     }
   return maxLength;
   }
   public static long choose(long total, long choose){
    if(total < choose)
        return 0;
    if(choose == 0 || choose == total)
        return 1;
    return choose(total-1,choose-1)+choose(total-1,choose);
}
    public static long bitwiseAdd(int n1, int n2) {
        int x = n1, y = n2;
        int xor, and, temp;
        and = x & y;
        xor = x ^ y;

        while (and != 0) {
            and <<= 1;
            temp = xor ^ and;
            and &= xor;
            xor = temp;
        }
        return (xor%2);
    }
    public static boolean compare(String a,String b){
        for (int i = 0; i < a.length(); i++) {
            if(a.charAt(i)!=b.charAt(i)&&a.charAt(i)!='?'){return false;}
        }
  return true;
    }*/
/*public static boolean isUniqueChars(String str) {
    // short circuit - supposed to imply that
    // there are no more than 256 different characters.
    // this is broken, because in Java, char's are Unicode,
    // and 2-byte values so there are 32768 values
    // (or so - technically not all 32768 are valid chars)
    if (str.length() > 256) {
        return false;
    }
    // checker is used as a bitmap to indicate which characters
    // have been seen already
    int checker = 0;
    for (int i = 0; i < str.length(); i++) {
        // set val to be the difference between the char at i and 'a'
        // unicode 'a' is 97
        // if you have an upper-case letter e.g. 'A' you will get a
        // negative 'val' which is illegal
        int val = str.charAt(i) - 'a';
        // if this lowercase letter has been seen before, then
        // the corresponding bit in checker will have been set and
        // we can exit immediately.
        if ((checker & (1 << val)) > 0) return false;
        // set the bit to indicate we have now seen the letter.
        checker |= (1 << val);
    }
    // none of the characters has been seen more than once.
    return true;
}*/
 /*private static int [] PossibilitiesAsBinaryTruthTable(int n) {
        int rows = (int) Math.pow(2,n);
        int [] possibleTrack=new int[rows];
        int trackPerRow;
        for (int i=0; i<rows; i++) {
            trackPerRow=0;
            for (int j=n-1; j>=0; j--) {
                if(((i/(int) Math.pow(2, j))%2)==0){trackPerRow--;}
                else if(((i/(int) Math.pow(2, j))%2)==1){trackPerRow++;}
            }
            possibleTrack[i]=trackPerRow;
        }
        return possibleTrack;
    }
    //Steins GCD Algo
     public static long gcd(long u, long v)
    {
        long shift;
        if (u == 0)
            return v;
        if (v == 0)
            return u;
        for (shift = 0; ((u | v) & 1) == 0; ++shift) 
        {
            u >>= 1;
            v >>= 1;
        }
        while ((u & 1) == 0)
            u >>= 1;
        do 
        {
            while ((v & 1) == 0)  
                v >>= 1;
 
            if (u > v) 
            {
                long t = v; 
                v = u; 
                u = t;
            }  
            v = v - u;                      
        } while (v != 0);
 
        return u << shift; 
    }*/
     public static long gcd(long p, long q) {
        if (q == 0) return p;
        else return gcd(q, p % q);
    }
    public static void main (String []  args) throws IOException 
    {
        Scanner s=new Scanner(System.in);
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));
        String line=br.readLine();
        String []nm=line.split(" ");
        int n=Integer.parseInt(nm[0]);int m=Integer.parseInt(nm[1]);
        String line2=br.readLine();
        String []nm2=line2.split(" ");
        int ix=Integer.parseInt(nm2[0]);int iy=Integer.parseInt(nm2[1]);;
        int k=Integer.parseInt(br.readLine());
        int x,y;
        boolean cond;
        long count=0;
        int temp=0;
        int tempx=0;
        int tempy=0;
        String line3;
        List<String>L=new ArrayList<>();
        //at each step we know for sure and it is a fact that the player is currently on one of the squares 
        //so the available squares is n-1,m-1 at any moment of time.
        for (int i = 0; i < k; i++) {
            line3=br.readLine();
            x=Integer.parseInt(line3.split(" ")[0]);y=Integer.parseInt(line3.split(" ")[1]);
            //the current pos is donated by ix,iy and the query is x[i],y[i]
            //check first if it is not out border
            if((ix+x<=n&&ix+x>=1)&&(iy+y<=m&&iy+y>=1)){
            //if we will be succeffly able to add at least one step of it then it is ok we can update the place 
            //now we will see how much update we will get by using division and modulus
            //we are calculating the number of steps with respect to orgin (1,1)
            if(x==0){temp=(m-iy)/Math.abs(y);if(y<0){temp=(iy-1)/Math.abs(y);}}
            if(y==0){temp=(n-ix)/Math.abs(x);if(x<0){temp=(ix-1)/Math.abs(x);}}
            if(x>0){tempx=(n-ix)/x;}
            if(x<0){tempx=(ix-1)/Math.abs(x);}
            if(y>0){tempy=(m-iy)/y;}
            if(y<0){tempy=(iy-1)/Math.abs(y);}
            if(x!=0&&y!=0){temp=Math.min(tempx, tempy);}
            //update direction
            ix+=(temp*x);iy+=(temp*y);
            //count number of steps taken
            count+=temp;
            }
           }
        
        System.out.print(count);
   }
}
   
