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.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.PriorityQueue;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {

        public long findMinAnswer(long[] bs, long as) {
            int l = 0;
            int r = bs.length;
            while (l + 1 < r) {
                int m = (l + r) >> 1;
                if (bs[m] < as) {
                    l = m;
                } else {
                    r = m;
                }
            }
            long res = Math.abs(bs[l] - as);
            if (r != bs.length) {
                res = Math.min(Math.abs(bs[r] - as), res);
            }
            return res;

        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int q = in.nextInt();
            long aSum = 0;
            for (int i = 0; i < n; i++) {
                if (i % 2 == 0) {
                    aSum += in.nextInt();
                } else {
                    aSum -= in.nextInt();
                }
            }
            int[] b = new int[m];
            for (int i = 0; m > i; i++)
                b[i] = in.nextInt();
            long[] bSum = new long[m - n + 1];
            for (int i = 0; i < n; i++) {
                if (i % 2 != 0) {
                    bSum[0] += b[i];
                } else {
                    bSum[0] -= b[i];
                }
            }
            for (int i = 1; i <= m - n; i++) {
                bSum[i] = bSum[i - 1];
                bSum[i] += b[i - 1];
                bSum[i] *= -1;
                if (n % 2 == 0) {
                    bSum[i] += b[i + n - 1];
                } else {
                    bSum[i] -= b[i + n - 1];
                }
            }
            long minB = Long.MAX_VALUE;
            for (int i = 0; i <= m - n; i++) {
                minB = Math.min(bSum[i], minB);
            }
            Arrays.sort(bSum);

            out.println(findMinAnswer(bSum, aSum * (-1)));
            for (int i = 0; i < q; i++) {
                int l = in.nextInt();
                int r = in.nextInt();
                int x = in.nextInt();
                if ((r - l) % 2 == 0) {
                    if (r % 2 == 0) {
                        aSum -= x;
                    } else {
                        aSum += x;
                    }
                }
                out.println(findMinAnswer(bSum, aSum * (-1)));
            }
        }
    }

    static class InputReader {
        public BufferedReader  reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextFloat() {
            return Double.parseDouble(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

    }
}
