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.
/**
 * @(#)Main862E.java
 *
 *
 * @author
 * @version 1.00 2017/12/5
 */
import java.math.*;
import java.io.*;
import java.util.*;

public class Main862E{
	static Scanner in;
	static PrintWriter out;

	static int n, m, q, ptr;
	static int[] a, b;

	static long origVal;
	static long[] partialSum, values;

	public static void main(String args[]){
		Scanner kbd = new Scanner(System.in);
		out = new PrintWriter(System.out);

		n = kbd.nextInt();
		m = kbd.nextInt();
		q = kbd.nextInt();

		a = new int[n + 1];
		b = new int[m + 1];
		partialSum = new long[m + 1];
		values = new long[m + 1];

		origVal = 0;
		for (int i = 1; i <= n; i++){
			a[i] = kbd.nextInt();
			if (i % 2 == 1){
				origVal += a[i];
			}
			else{
				origVal -= a[i];
			}
		}

		for (int i = 1; i <= m; i++){
			b[i] = kbd.nextInt();
			if (i % 2 == 1){
				partialSum[i] = partialSum[i - 1] + b[i];
			}
			else{
				partialSum[i] = partialSum[i - 1] - b[i];
			}
		}

		for (int j = 0; j <= m - n; j++){
			if (j % 2 == 1){
				values[j + 1] = (-1) * (partialSum[j + n] - partialSum[j]);
			}
			else{
				values[j + 1] = partialSum[j + n] - partialSum[j];
			}
		}

		ptr = m - n + 1;
		Arrays.sort(values, 1, ptr + 1);

		solve(origVal);
		for (int i = 1; i <= q; i++){
			int left, right, dx;
			left = kbd.nextInt();
			right = kbd.nextInt();
			dx = kbd.nextInt();
			if (right % 2 == 1){
				origVal += dx;
			}
			if (left % 2 == 0){
				origVal -= dx;
			}
			solve(origVal);
		}
		out.close();
	}

	public static void solve(long v){
		int left = 0, right = ptr + 1;
		while (right - left > 1){
			int mid = (left + right) / 2;
			if (values[mid] > v){
				right = mid;
			}
			else{
				left = mid;
			}
		}
		long res = 1;
		for (int i = 1; i < 62; i++) res *= 2;
		if (left > 0){
			res = Math.min(res, Math.abs(v - values[left]));
		}
		if (right <= ptr){
			res = Math.min(res, Math.abs(v - values[right]));
		}
		out.println(res);
		return;
	}
}