Dr. Evil is interested in math and functions, so he gave Mahmoud and Ehab array a of length n and array b of length m. He introduced a function f(j) which is defined for integers j, which satisfy 0 ≤ j ≤ m - n. Suppose, ci = ai - bi + j. Then f(j) = |c1 - c2 + c3 - c4... cn|. More formally, <image>. 

Dr. Evil wants Mahmoud and Ehab to calculate the minimum value of this function over all valid j. They found it a bit easy, so Dr. Evil made their task harder. He will give them q update queries. During each update they should add an integer xi to all elements in a in range [li;ri] i.e. they should add xi to ali, ali + 1, ... , ari and then they should calculate the minimum value of f(j) for all valid j.

Please help Mahmoud and Ehab.

Input

The first line contains three integers n, m and q (1 ≤ n ≤ m ≤ 105, 1 ≤ q ≤ 105) — number of elements in a, number of elements in b and number of queries, respectively.

The second line contains n integers a1, a2, ..., an. ( - 109 ≤ ai ≤ 109) — elements of a.

The third line contains m integers b1, b2, ..., bm. ( - 109 ≤ bi ≤ 109) — elements of b.

Then q lines follow describing the queries. Each of them contains three integers li ri xi (1 ≤ li ≤ ri ≤ n,  - 109 ≤ x ≤ 109) — range to be updated and added value.

Output

The first line should contain the minimum value of the function f before any update.

Then output q lines, the i-th of them should contain the minimum value of the function f after performing the i-th update .

Example

Input

5 6 3
1 2 3 4 5
1 2 3 4 5 6
1 1 10
1 1 -9
1 5 -1


Output

0
9
0
0

Note

For the first example before any updates it's optimal to choose j = 0, f(0) = |(1 - 1) - (2 - 2) + (3 - 3) - (4 - 4) + (5 - 5)| = |0| = 0.

After the first update a becomes {11, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(11 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6) = |9| = 9.

After the second update a becomes {2, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(2 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6)| = |0| = 0.

After the third update a becomes {1, 1, 2, 3, 4} and it's optimal to choose j = 0, f(0) = |(1 - 1) - (1 - 2) + (2 - 3) - (3 - 4) + (4 - 5)| = |0| = 0.
import java.io.* ;
import java.util.* ;
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*ModInverse(b))%mod ; }
	static long m(long a , long b){ return (a*b)%mod ; }
	static private long ModInverse(long a ){ return pow(a,mod-2) ; }
	static long pow(long a,long b){
		long val=a ; long ans=1 ;
		while(b!=0){
			if((b&1)==1) ans = m(ans,val) ; 
			val = m(val,val) ; 
			b>>=1 ;
		}
		return ans ;
	}	
}
class pair implements Comparable<pair>{
	int x ; int y ; int z ;
	pair(int x,int y,int z){ this.x=x ; this.y=y ; this.z=z ;}
	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) ;
	void Solve() throws IOException{
		int n = ip.i() ;
		int m = ip.i() ;
		int q = ip.i() ;
		long a[] = new long[n] ;
		long b[] = new long[m] ;
		for(int i=0 ; i<n ; i++) a[i] = ip.l() ;
		for(int i=0 ; i<m ; i++) b[i] = ip.l() ;
		long aval=0 ;
		int mul=1 ;
		for(int i=0 ; i<n ; i++){
			aval+=(mul*a[i]) ;
			mul=-mul ;
		}
		long bvl[] = new long[m-n+1] ;
		long bval = 0 ;
		mul = 1 ;
		for(int i=0 ; i<n ; i++){
			bval+=(mul*b[i]) ;
			mul=-mul ;
		}
		bvl[0]=bval ;
		for(int i=1 ; i<=(m-n) ; i++)
			bvl[i] = (b[i-1]-bvl[i-1])-(mul*b[i+n-1]) ;
		Arrays.sort(bvl) ;
		pln(findMinDist(bvl,aval,m,n)) ;
		while(q--!=0){
			int l = ip.i()-1 ;
			int r = ip.i()-1 ;
			int x = ip.i() ;
			if(((r-l)&1)==0)
				aval+=(((l&1)==0) ? x : -x) ;
			pln(findMinDist(bvl,aval,m,n)) ;
		}
	}
	long findMinDist(long bvl[],long aval,int m,int n){
		int idx = Arrays.binarySearch(bvl,aval) ;
		if(idx>=0)
			return 0 ;
		else{
			idx=-idx-1 ;
			long min = Long.MAX_VALUE ;
			if(idx>0) min = Long.min(min,abd(aval,bvl[idx-1])) ;
			if(idx<=(m-n)) min = Long.min(min,abd(aval,bvl[idx])) ;
			return min ;
		}
	}
	long abd(long x,long y){
		return x>y ? x-y : y-x ;
	}
	void p(Object o){
		op.print(o) ;
		op.flush() ;
	}
	void pln(Object o){
		op.println(o) ;
		op.flush() ;
	}
	void Finish(){
		op.flush();
		op.close();
	}	
}
@SuppressWarnings("serial")
class mylist extends ArrayList<pair>{
}
@SuppressWarnings("serial")
class myset extends TreeSet<Integer>{
	
}
@SuppressWarnings("serial")
class mystack extends Stack<pair>{
	
}
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; } 
} 