In Ann's favorite book shop are as many as n books on math and economics. Books are numbered from 1 to n. Each of them contains non-negative number of problems.

Today there is a sale: any subsegment of a segment from l to r can be bought at a fixed price. 

Ann decided that she wants to buy such non-empty subsegment that the sale operates on it and the number of math problems is greater than the number of economics problems exactly by k. Note that k may be positive, negative or zero.

Unfortunately, Ann is not sure on which segment the sale operates, but she has q assumptions. For each of them she wants to know the number of options to buy a subsegment satisfying the condition (because the time she spends on choosing depends on that).

Currently Ann is too busy solving other problems, she asks you for help. For each her assumption determine the number of subsegments of the given segment such that the number of math problems is greaten than the number of economics problems on that subsegment exactly by k.

Input

The first line contains two integers n and k (1 ≤ n ≤ 100 000,  - 109 ≤ k ≤ 109) — the number of books and the needed difference between the number of math problems and the number of economics problems.

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 2), where ti is 1 if the i-th book is on math or 2 if the i-th is on economics.

The third line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the number of problems in the i-th book.

The fourth line contains a single integer q (1 ≤ q ≤ 100 000) — the number of assumptions.

Each of the next q lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) describing the i-th Ann's assumption.

Output

Print q lines, in the i-th of them print the number of subsegments for the i-th Ann's assumption.

Examples

Input

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


Output

2
3
4
1


Input

4 0
1 2 1 2
0 0 0 0
1
1 4


Output

10

Note

In the first sample Ann can buy subsegments [1;1], [2;2], [3;3], [2;4] if they fall into the sales segment, because the number of math problems is greater by 1 on them that the number of economics problems. So we should count for each assumption the number of these subsegments that are subsegments of the given segment.

Segments [1;1] and [2;2] are subsegments of [1;2].

Segments [1;1], [2;2] and [3;3] are subsegments of [1;3].

Segments [1;1], [2;2], [3;3], [2;4] are subsegments of [1;4].

Segment [3;3] is subsegment of [3;4].
import java.io.* ;
import java.util.* ;
import java.util.stream.* ;
import java.text.* ;
import java.math.* ;
public class Codeshefcode{
	public static void main(String[] args) throws IOException{
		Solver Machine = new Solver() ;
		Machine.Solve() ;	
		Machine.Finish();
	}	
}
class Mod{
	static long mod=1000000007 ;
	static long d(long a,long b){ return (a*MI(b))%mod ; }
	static long m(long a , long b){ return (a*b)%mod ; }
	static private long MI(long a ){ return pow(a,mod-2) ; }
	static long pow(long a,long b){
		if(b<0) return pow(MI(a),-b) ;
		long val=a ; long ans=1 ;
		while(b!=0){
			if((b&1)==1) ans = (ans*val)%mod ; 
			val = (val*val)%mod ; 
			b>>=1 ;
		}
		return ans ;
	}	
}
class pair implements Comparable<pair>{
	int x ; int y ; 
	pair(int x,int y){ this.x=x ; this.y=y ;}
	public int compareTo(pair p){
		return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
	}
}
class Solver{	
	FasterScanner ip = new FasterScanner(System.in) ;	  
	PrintWriter op = new PrintWriter(System.out) ;
	int n,k,q,m,e,s,t[],km[],kp[],kz[] ;
	long a[],pre[],cnt[],res[],ans=0 ;
	int bsz ;
	class Query implements Comparable<Query>{
		int l,r,i ;
		Query(int l,int r,int i){
			this.l=l ; this.r=r ; this.i=i ;
		}
		public int compareTo(Query qr){
			return (this.l/bsz<qr.l/bsz ? -1 : (this.l/bsz>qr.l/bsz ? 1 : (this.r<qr.r ? -1 : (this.r>qr.r ? 1 : 0)))) ;		
		}
	}
	void compress(){
		myset st = new myset() ;
		for(int i=0 ; i<=n ; i++){
			st.add(pre[i]) ;
			st.add(pre[i]-k) ;
			st.add(pre[i]+k) ;
		}		
		mymap mp = new mymap() ;
		m=0 ;
		for(long elem : st) mp.put(elem,m++) ;
		for(int i=0 ; i<=n ; i++){
			km[i] = mp.get(pre[i]-k) ;
			kz[i] = mp.get(pre[i]) ;
			kp[i] = mp.get(pre[i]+k) ;
		}
		cnt = new long[m] ;
	}
	void Solve() throws IOException{
		n = ip.i() ; k = ip.i() ;
		bsz = (int) Math.sqrt(n) ;
		t = new int[n+1] ; km = new int[n+1] ; 
		kp = new int[n+1] ; kz = new int[n+1] ; 
		a = new long[n+1] ;	pre = new long[n+1] ; 
		for(int i=1 ; i<=n ; i++) t[i] = ip.i() ;
		for(int i=1 ; i<=n ; i++) a[i] = ip.l() ;
		for(int i=1 ; i<=n ; i++) a[i] = (t[i]==1 ? a[i] : -a[i]) ;
		for(int i=1 ; i<=n ; i++) pre[i] = pre[i-1]+a[i] ;
		compress();
		q = ip.i() ;
		Query qr[] = new Query[q] ;
		res = new long[q] ;
		for(int i=0 ; i<q ; i++){
			int l = ip.i() ; int r = ip.i() ;
			qr[i] = new Query(l-1,r,i) ;
		}
		Arrays.sort(qr) ;
		e=-1 ; s=0 ;
		for(int i=0 ; i<q ; i++){
			int l = qr[i].l,r = qr[i].r ;
			while(s>l) addl() ;			
			while(s<l) rmvl() ;
			while(e<r) addr() ;
			while(e>r) rmvr() ;
			res[qr[i].i]=ans ; 
		}
		for(int i=0 ; i<q ; i++) pln(res[i]) ;
	}
	void addl(){
		s-- ;
		ans+=cnt[kp[s]] ;
		cnt[kz[s]]++ ; 
	}
	void rmvl(){
		cnt[kz[s]]-- ;
		ans-=cnt[kp[s]] ;
		s++ ;	
	}	
	void addr(){
		e++ ;
		ans+=cnt[km[e]] ;
		cnt[kz[e]]++ ;
	}
	void rmvr(){
		cnt[kz[e]]-- ;
		ans-=cnt[km[e]] ;
		e-- ;
	}
	void p(Object o){
		op.print(o) ;
	}
	void pln(Object o){
		op.println(o) ;
	}
	void pArr(String name,Object arr,char type,int st){
		p(name+": [") ;
		if(type=='i'){ 
			int arr1[] = (int[]) arr ;
			for(int i=st ; i<arr1.length ; i++) p(arr1[i]+(i+1==arr1.length ? "" : ",")) ;		 
		}else if(type=='l'){ 
			long arr1[] = (long[]) arr ;
			for(int i=st ; i<arr1.length ; i++) p(arr1[i]+(i+1==arr1.length ? "" : ",")) ;
		}else if(type=='b'){
			boolean arr1[] = (boolean[]) arr ;
			for(int i=0 ; i<arr1.length ; i++) p((arr1[i] ? 1 : 0)+(i+1==arr1.length ? "" : ",")) ;
		}
		pln("]") ;
	}	
	void Finish(){
		op.flush();
		op.close();
	}	
}
@SuppressWarnings("serial")
class mylist extends ArrayList<Integer>{
}
@SuppressWarnings("serial")
class myset extends TreeSet<Long>{
}
@SuppressWarnings("serial")
class mystack extends Stack<Integer>{
}
@SuppressWarnings("serial")
class mymap extends TreeMap<Long,Integer>{
}
class FasterScanner { 
	private InputStream mIs;
	private byte[] buf = new byte[1024];
	private int curChar,numChars;
	public FasterScanner() { this(System.in); } 
	public FasterScanner(InputStream is) { mIs = is;} 
	public int read() {
		if (numChars == -1) throw new InputMismatchException();    
		if (curChar >= numChars) {
			curChar = 0;
			try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}
			if (numChars <= 0) return -1;            
		}
		return buf[curChar++];
	} 
	public String nextLine(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString();
	} 
	public String S(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		}while (!isSpaceChar(c));
		return res.toString();
	} 
	public long l(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		long res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read();
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public int i(){
		int c = read() ;
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		int res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
		res *= 10 ; res += c - '0' ; c = read() ;
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public int[] ia(int n){
		int a[] = new int[n] ;
		for(int i=0 ; i<n ; i++) a[i] = i() ;
		return a ;
	}
	public long[] la(int n){
		long a[] = new long[n] ;
		for(int i=0 ; i<n ; i++) a[i] = l() ;
		return a ;
	}
	public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
	public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
}