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.ArrayDeque;
import java.util.Deque;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class randoms {

    static int[] neat = new int[]{1, -1};

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        StringTokenizer st = new StringTokenizer(br.readLine());
        int N = Integer.parseInt(st.nextToken());
        int M = Integer.parseInt(st.nextToken());
        int Q = Integer.parseInt(st.nextToken());
        long a = 0;
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            a += Integer.parseInt(st.nextToken()) * neat[i % 2];
        }
        TreeSet<Long> all = new TreeSet<>();
        Deque<Integer> dq = new ArrayDeque();
        long b = 0;
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < M; i++) {
            int c = Integer.parseInt(st.nextToken()) * neat[i % 2];
            dq.addLast(c);
            b += c;
            if (dq.size() == N + 1) {
                b -= dq.pollFirst();
            }
            if (dq.size() == N) {
                all.add(b * neat[(i - N + 1) % 2]);
            }
        }
//        System.out.println(all.size());
//        System.out.println(a);
//        for(long i : all) System.out.println(i);
        pw.println(find(all, a));
        for (int i = 0; i < Q; i++) {
            st = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());
            if ((r - l) % 2 == 0) {
                l %= 2;
                l ^= 1;
                a += neat[l] * Integer.parseInt(st.nextToken());
            }
            pw.println(find(all, a));
        }
        pw.close();
        br.close();
    }

    static long find(TreeSet<Long> set, long val) {
        Long up = set.ceiling(val);
        Long down = set.floor(val);
        if (up == null) {
            return val - down;
        }
        if (down == null) {
            return up - val;
        }
        return Math.min(val - down, up - val);
    }
}
