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.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;

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

    static class TaskE {
        public void solve(int testNumber, FastScanner in, FastPrinter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int q = in.nextInt();
            long[] a = in.nextLongArray(n);
            long[] b = in.nextLongArray(m);
            long sa = 0;
            for (int i = 0; i < n; i++) {
                sa = i % 2 == 0 ? sa + a[i] : sa - a[i];
            }
            TreeSet<Long> sums = new TreeSet<>();
            long sb = 0;
            for (int i = 0; i < n; i++) {
                sb = i % 2 == 0 ? sb + b[i] : sb - b[i];
            }
            sums.add(sb);
            for (int i = 0; i < m - n; i++) {
                sb = -sb + b[i];
                sb = n % 2 == 0 ? sb - b[n + i] : sb + b[n + i];
                sums.add(sb);
            }
            out.printLine(find(sa, sums));
            for (int i = 0; i < q; i++) {
                int l = in.nextInt() - 1;
                int r = in.nextInt();
                int x = in.nextInt();
                if ((r - l) % 2 == 1) {
                    sa = l % 2 == 0 ? sa + x : sa - x;
                }
                out.printLine(find(sa, sums));
            }
        }

        long find(long sa, TreeSet<Long> sums) {
            if (sums.lower(sa) == null) return sums.first() - sa;
            if (sums.higher(sa) == null) return sa - sums.last();
            return Math.min(sa - sums.lower(sa), sums.higher(sa) - sa);
        }

    }

    static class FastPrinter {
        private final PrintWriter writer;

        public FastPrinter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public FastPrinter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void close() {
            writer.close();
        }

        public void printLine(long i) {
            writer.println(i);
        }

    }

    static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            this.reader = new BufferedReader(new InputStreamReader(stream));
            this.st = new StringTokenizer("");
        }

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

        public long nextLong() {
            return Long.parseLong(next());
        }

        public String next() {
            while (!st.hasMoreTokens()) {
                st = new StringTokenizer(readLine());
            }
            return st.nextToken();
        }

        private String readLine() {
            String line = tryReadLine();
            if (line == null) throw new InputMismatchException();
            return line;
        }

        private String tryReadLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }

        public long[] nextLongArray(int size) {
            long[] array = new long[size];
            for (int i = 0; i < size; i++)
                array[i] = nextLong();
            return array;
        }

    }
}

