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.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
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);
        EMahmoudAndEhabAndTheFunction solver = new EMahmoudAndEhabAndTheFunction();
        solver.solve(1, in, out);
        out.close();
    }

    static class EMahmoudAndEhabAndTheFunction {
        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();
            long[] a = new long[n];
            long[] b = new long[m];
            for (int i = 0; i < n; i++) a[i] = sc.nextInt();
            for (int i = 0; i < m; i++) b[i] = sc.nextInt();
            long cons = 0;
            for (int i = 0; i < n; i++) {
                cons += a[i] * (i % 2 == 0 ? 1 : -1);
            }
            TreeSet<Long> ts = new TreeSet<>();
            long[] sumb = new long[m];
            sumb[0] = -b[0];
            for (int i = 1; i < m; i++) sumb[i] = sumb[i - 1] + b[i] * (i % 2 == 0 ? -1 : 1);
            // pw.println(Arrays.toString(sumb));
            for (int i = 0; i <= m - n; i++) {
                long tmp = sumb[i + n - 1] - ((i - 1 >= 0) ? sumb[i - 1] : 0);
                if (i % 2 == 0) ts.add(tmp);
                else ts.add(-tmp);
            }

            pw.println(Math.min(Math.abs(cons + (ts.ceiling(-cons) == null ? Long.MAX_VALUE : ts.ceiling(-cons))), Math.abs(cons + (ts.floor(-cons) == null ? Long.MAX_VALUE : ts.floor(-cons)))));
            for (int i = 0; i < q; i++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                long x = sc.nextInt();
                if ((r - l + 1) % 2 == 0) {
                    pw.println(Math.min(Math.abs(cons + (ts.ceiling(-cons) == null ? Long.MAX_VALUE : ts.ceiling(-cons))), Math.abs(cons + (ts.floor(-cons) == null ? Long.MAX_VALUE : ts.floor(-cons)))));
                } else {
                    if (l % 2 == 0) cons -= x;
                    else cons += x;
                    pw.println(Math.min(Math.abs(cons + (ts.ceiling(-cons) == null ? Long.MAX_VALUE : ts.ceiling(-cons))), Math.abs(cons + (ts.floor(-cons) == null ? Long.MAX_VALUE : ts.floor(-cons)))));
                }

            }
            pw.flush();
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

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

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

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

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

    }
}

