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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

import static java.lang.Math.random;

/**
 * @author Don Li
 */
public class MahmoudEhabFunction {
    void solve() {
        int n = in.nextInt(), m = in.nextInt(), q = in.nextInt();
        int[] a = new int[n], b = new int[m];
        for (int i = 0; i < n; i++) a[i] = in.nextInt();
        for (int i = 0; i < m; i++) b[i] = in.nextInt();
        
        long sa = 0;
        for (int i = 0; i < n; i++) {
            if ((i & 1) == 0) sa += a[i];
            else sa -= a[i];
        }
        
        long[] sb = new long[m - n + 1];
        for (int i = 0; i < n; i++) {
            if ((i & 1) == 0) sb[0] += b[i];
            else sb[0] -= b[i];
        }
        for (int i = 1; i <= m - n; i++) {
            sb[i] = b[i - 1] - sb[i - 1];
            if ((n & 1) == 0) sb[i] -= b[n + i - 1];
            else sb[i] += b[n + i - 1];
        }
        
        sort(sb);
        
        out.println(findMin(sb, sa));
        for (int i = 0; i < q; i++) {
            int l = in.nextInt() - 1, r = in.nextInt(), x = in.nextInt();
            if (((r - l) & 1) > 0) {
                if ((l & 1) == 0) sa += x;
                else sa -= x;
            }
            out.println(findMin(sb, sa));
        }
    }
    
    long findMin(long[] a, long k) {
        int i = lowerBound(a, k);
        long min = Long.MAX_VALUE;
        if (i < a.length) min = Math.min(min, Math.abs(k - a[i]));
        if (i - 1 >= 0) min = Math.min(min, Math.abs(k - a[i - 1]));
        return min;
    }
    
    int lowerBound(long[] a, long k) {
        int lb = -1, ub = a.length;
        while (ub - lb > 1) {
            int m = (lb + ub) >> 1;
            if (a[m] >= k) ub = m;
            else lb = m;
        }
        return ub;
    }
    
    void sort(long[] a) {
        int n = a.length;
        for (int i = 0; i < n; i++) {
            int j = (int) (i + random() * (n - i));
            long tmp = a[i];
            a[i] = a[j];
            a[j] = tmp;
        }
        Arrays.sort(a);
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new MahmoudEhabFunction().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
