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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;

/**
 * Created by henryhuypham on 9/20/17.
 */
public class Main {

    private static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    private static PrintWriter out = new PrintWriter(System.out);

    private static String[] nextLine() throws IOException {
        return in.readLine().split(" ");
    }

    private static int ivl(String val) {
        return Integer.parseInt(val);
    }

    public static void main(String[] args) throws IOException {
        String[] inp = nextLine();
        int n = ivl(inp[0]), m = ivl(inp[1]), q = ivl(inp[2]);
        long[] A = new long[n + 1], B = new long[m + 1];

        inp = nextLine();
        for (int i = 1; i <= n; i++) A[i] = ivl(inp[i - 1]);
        inp = nextLine();
        for (int i = 1; i <= m; i++) B[i] = ivl(inp[i - 1]);

        long sa = 0;
        long[] sb = new long[m - n + 1];
        for (int i = 1; i <= n; i++) {
            sa += A[i] * (i % 2 == 1 ? 1 : -1);
            sb[0] += B[i] * (i % 2 == 1 ? 1 : -1);
        }
        for (int i = 2; i <= sb.length; i++) {
            sb[i - 1] = -sb[i - 2] + B[i - 1] - B[i + n - 1] * ((n - 1) % 2 == 1 ? 1 : -1);
        }
//        for (int i = 0; i < sb.length; i++) System.out.println(sb[i]);

        Arrays.sort(sb);

        for (int i = 0; i <= q; i++) {
            if (i != 0) {
                inp = nextLine();
            } else {
                inp = new String[]{"1", "" + n, "0"};
            }
            int l = ivl(inp[0]), r = ivl(inp[1]), x = ivl(inp[2]);
            if ((r - l) % 2 == 0) {
                sa += l % 2 == 1 ? x : -x;
            }

            int left = 0, right = sb.length - 1;
            int mid = -1;
            while (left <= right) {
                mid = (left + right) / 2;
                if (sa < sb[mid]) {
                    right = mid - 1;
                } else if (sa > sb[mid]) {
                    left = mid + 1;
                } else {
                    break;
                }
            }
            long res = Math.abs(sb[0] - sa);
            for (int j = -1; j <= 1; j++) {
                int idx = mid + j;
                if (idx >=0 && idx < sb.length) {
                    res = Math.min(res, Math.abs(sb[idx] - sa));
                }
            }
            out.println(res);
        }
        out.flush();
    }
}
