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;
long long int pow(int x, int y) {
  if (y % 2 == 0)
    return 1;
  else
    return -1;
}
long long absll(long long x) {
  if (x > 0) return x;
  return -x;
}
long long an(long long x, set<long long int>& s) {
  set<long long>::iterator it = s.lower_bound(x);
  if (it == s.end()) it--;
  long long ret = absll((*it) - x);
  if (it != s.begin()) it--;
  return min(ret, absll((*it) - x));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, sum = 0, count = 0, m, flag = 0, ans = 0, k;
  cin >> n >> m >> k;
  vector<long long int> a(n), b(m);
  set<long long int> s;
  for (int i = 0; i < n; i++) cin >> a[i], sum += a[i] * pow(-1, i);
  for (int i = 0; i < m; i++) cin >> b[i], b[i] = b[i] * pow(-1, i);
  for (int i = 0; i < m; i++) {
    ans += b[i];
    if (i >= n) {
      if ((i - n) % 2 == 0)
        s.insert(ans - b[i]);
      else
        s.insert((ans - b[i]) * -1);
      ans -= b[i - n];
    }
  }
  if ((m - n) % 2 == 0)
    s.insert(ans);
  else
    s.insert(-ans);
  cout << an(sum, s) << endl;
  while (k--) {
    long long int l, r, x;
    cin >> l >> r >> x;
    if (abs(r - l + 1) % 2 != 0) {
      sum += x * pow(-1, r - 1);
    }
    cout << an(sum, s) << endl;
  }
  return 0;
}
