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;
const int mod = 1000000007;
inline int ri() {
  int res = 0, neg = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') neg = 1, c = getchar();
  while (c >= '0' && c <= '9') {
    res = (res << 1) + (res << 3) + c - 48;
    c = getchar();
  }
  return neg ? -res : res;
}
inline long long int rl() {
  long long int res = 0, neg = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') neg = 1, c = getchar();
  while (c >= '0' && c <= '9') {
    res = (res << 1) + (res << 3) + c - 48;
    c = getchar();
  }
  return neg ? -res : res;
}
long long int calc(vector<long long int> const &a, long long int const key) {
  long long int beg = 0, las = a.size() - 1;
  while (beg <= las) {
    long long int mid = beg + (las - beg) / 2;
    if (a[mid] > key)
      las = mid - 1;
    else if (a[mid] < key)
      beg = mid + 1;
    else
      return 0;
  }
  if (las < 0) return a[beg] - key;
  if (beg >= a.size()) return key - a[las];
  return min(key - a[las], a[beg] - key);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n = rl(), m = rl(), q = rl();
  long long int a[n], b[m];
  for (int i = 0; i < n; ++i) a[i] = rl();
  for (int i = 0; i < m; ++i) b[i] = rl();
  long long int i = 0, s = 0, p = 0;
  for (i = 0; i < n; ++i) {
    if (i % 2)
      p -= a[i], s -= b[i];
    else
      p += a[i], s += b[i];
  }
  vector<long long int> bsum;
  bsum.push_back(s);
  long long int k;
  if (n % 2)
    k = 1;
  else
    k = -1;
  for (; i < m; ++i) {
    s -= (k * b[i] + b[i - n]);
    s = -s;
    bsum.push_back(s);
  }
  sort(bsum.begin(), bsum.end());
  long long int l, r, x, tmp = 0;
  cout << calc(bsum, p) << '\n';
  while (q--) {
    l = rl(), r = rl(), x = rl();
    --l, --r;
    if ((r - l) % 2)
      tmp = 0;
    else if (l % 2)
      tmp = -x;
    else
      tmp = x;
    p += tmp;
    cout << calc(bsum, p) << '\n';
  }
  return 0;
}
