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 CF862_D2_E {
	public static void main(String[] args)throws Throwable {
		MyScanner sc=new MyScanner();
		PrintWriter pw=new PrintWriter(System.out);
		
		int n=sc.nextInt();
		int m=sc.nextInt();
		int q=sc.nextInt();
		long c=0;
		int [] a=new int[n];
		int [] b=new int [m];
		for(int i=0;i<n;i++){
			a[i]=sc.nextInt();
			if(i%2==0)
				c+=a[i];
			else
				c-=a[i];
		}
		dif=m-n+1;
		sum2=new long [dif];
		for(int i=0;i<m;i++){
			b[i]=sc.nextInt();
			if(i<n){
				if(i%2==0)
					sum2[0]-=b[i];
				else
					sum2[0]+=b[i];
			}
		}
		for(int i=1;i<dif;i++){
			sum2[i]=-1*sum2[i-1]-b[i-1];
			if(n%2==0){
				sum2[i]+=b[i+n-1];
			}else{
				sum2[i]-=b[i+n-1];
			}
		}
		Arrays.sort(sum2);
		pw.println(solve(c));
		while(q-->0){
			int l=sc.nextInt()-1;
			int r=sc.nextInt()-1;
			int x=sc.nextInt();
			int d=r-l+1;
			if(d%2==1){
				if(l%2==0)
					c+=x;
				else
					c-=x;
			}
			pw.println(solve(c));
		}
		
		pw.flush();
		pw.close();
	}
	static int dif;
	static long [] sum2;
	
	static long solve(long c){
		long ans=Long.MAX_VALUE;
		int start=0,end=dif-1;
		while(start<=end){
			int mid=((start+end)>>1);
			ans=Math.min(ans, Math.abs(c+sum2[mid]));
			if(sum2[mid]>-c)
				end=mid-1;
			else
				start=mid+1;
		}
		return ans;
	}
	
	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {while (st == null || !st.hasMoreElements()) {
			try {st = new StringTokenizer(br.readLine());}
			catch (IOException e) {e.printStackTrace();}}
		return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String nextLine(){String str = "";
		try {str = br.readLine();}
		catch (IOException e) {e.printStackTrace();}
		return str;}
	}
}