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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.FileReader;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author OmarYasser
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        static ArrayList<Long> bValues = new ArrayList<>();

        public void solve(int testNumber, Scanner sc, PrintWriter out) {
            int n = sc.nextInt(), m = sc.nextInt(), q = sc.nextInt();
            int[] a = new int[n], b = new int[m];
            for (int i = 0; i < n; i++)
                a[i] = sc.nextInt();
            for (int i = 0; i < m; i++)
                b[i] = sc.nextInt();

            long sumA = 0;
            for (int i = 0; i < n; i++)
                if (i % 2 == 0)
                    sumA += a[i];
                else
                    sumA -= a[i];

            long sumB = 0;
            for (int i = 0; i < n; i++)
                if (i % 2 == 0)
                    sumB -= b[i];
                else
                    sumB += b[i];
            bValues.add(sumB);
            for (int i = 1; i < m; i++) {
                if (i + n - 1 >= m) break;
                sumB += b[i - 1];

                sumB *= -1;
                if ((i + n - 1 + i) % 2 == 0)
                    sumB -= b[i + n - 1];
                else
                    sumB += b[i + n - 1];
                bValues.add(sumB);
            }

            Collections.sort(bValues);
            out.println(Math.abs(sumA + best(-sumA)));
            while (q-- > 0) {
                int l = sc.nextInt(), r = sc.nextInt(), val = sc.nextInt();
                int len = r - l + 1;
                if (len % 2 == 1) {
                    if (l % 2 == 0) sumA -= val;
                    else sumA += val;
                }
                out.println(Math.abs(sumA + best(-sumA)));
            }
        }

        static long best(long val) {
            int lo = 0, hi = bValues.size() - 1;
            long best = bValues.get(0);
            for (int cnt = 0; cnt <= 20; cnt++) {
                int mid = lo + ((hi - lo) >> 1);
                best = Math.abs(val - bValues.get(mid)) < Math.abs(val - best) ? bValues.get(mid) : best;
                if (val > bValues.get(mid)) {
                    lo = Math.min(hi, mid + 1);
                } else {
                    hi = Math.max(lo, mid - 1);
                }
            }
            return best;
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader f) {
            br = new BufferedReader(f);
        }

        public String next() {
            while (st == null || !st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

