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.
#include <bits/stdc++.h>
using namespace std;
inline long long Readint() {
  long long ret = 0;
  int f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f *= -1;
  } while (ch >= 0 && (ch < '0' || ch > '9'));
  while ('0' <= ch && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * f;
}
void open() {
  freopen("E.in", "r", stdin);
  freopen("E.out", "w", stdout);
}
void close() {
  fclose(stdin);
  fclose(stdout);
}
const int MAXN = 101010;
int n, m, q;
long long sb[MAXN];
int a[MAXN];
int b[MAXN];
long long A;
int t;
void init() {
  scanf("%d", &n), scanf("%d", &m), scanf("%d", &q);
  for (int i = (1), i_END_ = (n); i <= i_END_; i++) scanf("%d", &a[i]);
  for (int i = (1), i_END_ = (m); i <= i_END_; i++) scanf("%d", &b[i]);
  for (int i = (1), i_END_ = (n); i <= i_END_; i++)
    if (i & 1)
      sb[0] += b[i];
    else
      sb[0] -= b[i];
  for (int i = (n + 1), i_END_ = (m); i <= i_END_; i++) {
    sb[i - n] = sb[i - n - 1] - b[i - n];
    sb[i - n] = -sb[i - n];
    if (n & 1)
      sb[i - n] += b[i];
    else
      sb[i - n] -= b[i];
  }
  t = m - n;
  sort(sb, sb + 1 + t);
  t = unique(sb, sb + 1 + t) - sb - 1;
  for (int i = (1), i_END_ = (n); i <= i_END_; i++)
    if (i & 1)
      A += a[i];
    else
      A -= a[i];
}
long long check(int l, int r, long long num) {
  if (l == r) return llabs(sb[l] - num);
  if (l + 1 == r) return min(llabs(sb[l] - num), llabs(sb[r] - num));
  int mid = (l + r) / 2;
  if (sb[mid] < num)
    return check(mid, r, num);
  else if (num < sb[mid])
    return check(l, mid, num);
  else
    return 0;
}
void work() {
  int x, y, d;
  long long T;
  cout << check(0, t, A) << endl;
  for (int i = (1), i_END_ = (q); i <= i_END_; i++) {
    scanf("%d", &x), scanf("%d", &y), scanf("%d", &d);
    if ((y - x + 1) % 2 == 0)
      T = A;
    else if (x & 1)
      T = A + d;
    else
      T = A - d;
    cout << check(0, t, T) << endl;
    A = T;
  }
}
int main() {
  int _ = 0;
  init();
  work();
  close();
  return 0;
}
