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.lang.*;
import java.util.*;

public class Main {
  public static void main(String args[]) { new Main().out.flush(); }
  PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out), false);
  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  StringTokenizer st = null;
  String nxttok() {
    try { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); }
    catch (IOException e) { throw new RuntimeException(e); }
    return st.nextToken();
  }
  int nxtint() { return Integer.parseInt(nxttok()); }
  long nxtlong() { return Long.parseLong(nxttok()); }
  double nxtdouble() { return Double.parseDouble(nxttok()); }

  /////////////////////////////////////////////////////////////

  static int maxn = 101010;

  int n, m, q;
  int a[] = new int[maxn], b[] = new int[maxn];

  long getsign(int index) { return index % 2 == 1 ? -1 : 1; }

  Long s[];
  long findans(long as) {
    int pos = Arrays.binarySearch(s, as);
    if (pos < 0) pos = -pos;
    long ans = Long.MAX_VALUE;
    for (int i = Math.max(0, pos - 2); i < Math.min(pos + 3, s.length); ++i) {
      ans = Math.min(ans, Math.abs(as - s[i]));
    }
    return ans;
  }

  Main() {
    n = nxtint(); m = nxtint(); q = nxtint();
    for (int i = -1; ++i < n; ) a[i] = nxtint();
    for (int i = -1; ++i < m; ) b[i] = nxtint();

    long bs = 0;
    for (int i = -1; ++ i < n - 1; ) bs += b[i] * getsign(i);
    s = new Long[m - n + 1];
    for (int i = n - 2, f = 0; ++i < m; ++f) {
      bs += b[i] * getsign(i);
      s[f] = bs * getsign(f);
      bs -= b[f] * getsign(f);
      //System.err.println(f + " " + s[f]);
    }
    Arrays.sort(s);

    long as = 0;
    for (int i = -1; ++i < n; ) as += a[i] * getsign(i);
    out.println(findans(as));
    while (q-- > 0) {
      int u = nxtint() - 1, v = nxtint() - 1, x = nxtint();
      if ((u - v) % 2 == 0) {
        as += x * getsign(u);
      }
      out.println(findans(as));
    }
  }
}
