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 a[100001], b[100001], av = 0, n, m, q;
long long bj[100001], nbj[100001];
void pm() {
  long long* ba = lower_bound(bj, bj + m - n + 1, -av);
  long long* bb = lower_bound(nbj, nbj + m - n + 1, av);
  (0);
  long long res;
  if (ba == bj + m - n + 1)
    res = abs(av - *bb), (0);
  else if (bb == nbj + m - n + 1)
    res = abs(av + *ba), (0);
  else
    res = min(abs(av - *bb), abs(av + *ba)), (0);
  cout << res << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  bool ls;
  for (int i = 1; i <= n; i++) {
    if (i % 2)
      bj[0] -= b[i], ls = false;
    else
      bj[0] += b[i], ls = true;
  }
  for (int j = 1; j <= m - n; j++) {
    bj[j] = -bj[j - 1] - b[j] + (ls ? 1 : -1) * b[j + n];
  }
  for (int j = 0; j <= m - n; j++) nbj[j] = -bj[j], (0);
  sort(bj, bj + m - n + 1);
  sort(nbj, nbj + m - n + 1);
  for (int i = 1; i <= n; i++) {
    if (i % 2)
      av += a[i];
    else
      av -= a[i];
  }
  pm();
  while (q--) {
    long long l, r, x;
    cin >> l >> r >> x;
    if (l % 2 == r % 2) {
      if (l % 2)
        av += x;
      else
        av -= x;
    }
    pm();
  }
}
