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.util.*;
import java.io.*;
 
public class Solution {
	static TreeMap<Long, Long> multiset = new TreeMap<Long, Long>();
	
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer token = new StringTokenizer(in.readLine());
		int n = Integer.parseInt(token.nextToken());
		int m = Integer.parseInt(token.nextToken());
		int q = Integer.parseInt(token.nextToken());
		token = new StringTokenizer(in.readLine());
		long[] A = new long[n];
		for(int a = 0; a < n; a++)
			A[a] = Long.parseLong(token.nextToken());
		token = new StringTokenizer(in.readLine());
		long[] B = new long[m];
		for(int a = 0; a < m; a++)
			B[a] = Long.parseLong(token.nextToken());
		long AS = 0;
		for(int a = 0; a < n; a++)
			AS += A[a] * (a % 2 == 0 ? 1 : -1);
		long[] dp = new long[m - n + 1];
		for(int a = 0; a < n; a++)
			dp[0] += (-B[a]) * (a % 2 == 0 ? 1 : -1);
		for(int a = n; a < m; a++) {
			dp[a - n + 1] = -(dp[a - n] - (-B[a - n]))
				+ (-B[a]) * ((n - 1) % 2 == 0 ? 1 : -1);
		}
		for(int a = 0; a < m - n + 1; a++)
			add(dp[a]);
		Long less = multiset.floorKey(-AS);
		if(less == null)
			less = Long.MIN_VALUE / 2;
		Long more = multiset.ceilingKey(-AS);
		if(more == null)
			more = Long.MAX_VALUE / 2;
		System.out.println(Math.min(-(less + AS), more + AS));
		for(int a = 0; a < q; a++) {
			token = new StringTokenizer(in.readLine());
			int l = Integer.parseInt(token.nextToken());
			int r = Integer.parseInt(token.nextToken());
			long x = Long.parseLong(token.nextToken());
			l--;
			r--;
			if((r - l + 1) % 2 == 1) {
				if(l % 2 == 0)
					AS += x;
				else
					AS -= x;
			}
			less = multiset.floorKey(-AS);
			if(less == null)
				less = Long.MIN_VALUE / 2;
			more = multiset.ceilingKey(-AS);
			if(more == null)
				more = Long.MAX_VALUE / 2;
			System.out.println(Math.min(-(less + AS), more + AS));
		}
	}
	
	static void add(long x){
		if(multiset.containsKey(x)){
			multiset.put(x, multiset.get(x) + 1);
		} else {
			multiset.put(x, 1L);
		}
	}

	static void remove(long x){
		multiset.put(x, multiset.get(x) - 1);
		if(multiset.get(x) == 0){
			multiset.remove(x);
		}
	}
}