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;
ifstream fin("split.in");
ofstream fout("split.out");
const long long N = 1e5, INF = 1e18, MOD = 1e9 + 7;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n, m, q;
  cin >> n >> m >> q;
  vector<long long> sums(m + 1), odd(m + 1), even(m + 1), newsums;
  long long sumA = 0;
  for (int i = 1; i <= n; i++) {
    long long first;
    cin >> first;
    if (i % 2 == 1)
      sumA += first;
    else
      sumA -= first;
  }
  for (int i = 1; i <= m; i++) {
    long long first;
    cin >> first;
    if (i % 2 == 1) {
      if (i >= 2)
        odd[i] = odd[i - 2] + first;
      else
        odd[i] = first;
    } else {
      if (i >= 2)
        even[i] = even[i - 2] + first;
      else
        even[i] = first;
    }
    if (i >= n) {
      long long suma = 0;
      if ((i - n + 1) % 2 == 1) {
        if (i > n)
          suma -= odd[i - !(i % 2)] - odd[i - n - 1];
        else
          suma -= odd[i - !(i % 2)];
        suma += even[i - i % 2] - even[i - n];
      } else {
        suma += odd[i - !(i % 2)] - odd[i - n];
        if (i > n)
          suma -= even[i - i % 2] - even[i - n - 1];
        else
          suma -= even[i - i % 2];
      }
      sums[i] = suma;
    }
  }
  for (int i = 1; i <= m - n + 1; i++) {
    newsums.push_back(sums[n + i - 1]);
  }
  sort(newsums.begin(), newsums.end());
  vector<long long>::iterator it =
      lower_bound(newsums.begin(), newsums.end(), -sumA);
  if (it == sums.end()) it--;
  long long ans = abs(sumA + *it);
  if (it != newsums.begin()) it--;
  ans = min(ans, abs(sumA + *it));
  cout << ans << '\n';
  for (int p = 1; p <= q; p++) {
    long long l, r, first;
    cin >> l >> r >> first;
    if ((r - l + 1) % 2 == 1) {
      if (l % 2 == 1)
        sumA += first;
      else
        sumA -= first;
    }
    vector<long long>::iterator it =
        lower_bound(newsums.begin(), newsums.end(), -sumA);
    if (it == sums.end()) it--;
    ans = abs(sumA + *it);
    if (it != newsums.begin()) it--;
    ans = min(ans, abs(sumA + *it));
    cout << ans << '\n';
  }
}
