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;
vector<long long> cand;
long long solve(long long goal) {
  int indx = 0;
  int minim = 0, maxim = ((int)(cand).size()) - 1;
  while (minim <= maxim) {
    int mid = (minim + maxim) / 2;
    if (cand[mid] <= goal) {
      minim = mid + 1;
      indx = max(indx, mid);
    } else {
      maxim = mid - 1;
    }
  }
  long long ans = abs(cand[indx++] - goal);
  if (indx < ((int)(cand).size())) ans = min(ans, cand[indx] - goal);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int N, M, Q;
  cin >> N >> M >> Q;
  long long a_sum = 0, s = 1;
  for (int i = (0); i < (N); i++) {
    int x;
    cin >> x;
    a_sum += s * x;
    s = -s;
  }
  vector<long long> b(M);
  for (int i = (0); i < (M); i++) {
    cin >> b[i];
  }
  long long cur = 0;
  s = 1;
  for (int i = (0); i < (N); i++) {
    cur += s * b[i];
    s = -s;
  }
  s = (N % 2) == 1 ? 1 : -1;
  cand.push_back(cur);
  for (int i = (N); i < (M); i++) {
    cur -= b[i - N];
    cur = -cur;
    cur += b[i] * s;
    cand.push_back(cur);
  }
  sort(cand.begin(), cand.end());
  cout << solve(a_sum) << "\n";
  for (int q = (0); q < (Q); q++) {
    int l, r, x;
    cin >> l >> r >> x;
    int amnt = r - l + 1;
    if (amnt % 2 == 1) {
      if (r % 2 == 0)
        a_sum -= x;
      else
        a_sum += x;
    }
    long long ans = solve(a_sum);
    cout << ans << "\n";
  }
  return 0;
}
