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 ceil(long long a, long long b) {
  long long res = a / b;
  if (a % b != 0) res++;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> a(n), b(m);
  long long total = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 0) {
      total += a[i];
    } else {
      total -= a[i];
    }
  }
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  set<long long> pos_sum;
  long long sum_even = 0, sum_odd = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      sum_even += b[i];
    } else {
      sum_odd += b[i];
    }
  }
  for (int j = 0; j < m - n + 1; j++) {
    if (j != 0) {
      if ((j - 1) % 2 == 0) {
        sum_even -= b[j - 1];
      } else {
        sum_odd -= b[j - 1];
      }
      if ((j + n - 1) % 2 == 0) {
        sum_even += b[j + n - 1];
      } else {
        sum_odd += b[j + n - 1];
      }
    }
    if (j % 2 == 0) {
      pos_sum.insert(-sum_odd + sum_even);
    } else {
      pos_sum.insert(-sum_even + sum_odd);
    }
  }
  auto find = [&]() {
    auto ub = pos_sum.upper_bound(total);
    long long res = LONG_LONG_MAX;
    if (ub != pos_sum.end()) {
      res = min(res, abs(total - *ub));
    }
    if (ub != pos_sum.begin()) {
      ub--;
      res = min(res, abs(total - *ub));
    }
    return res;
  };
  cout << find() << '\n';
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    l--, r--;
    long long len = r - l + 1;
    if (l % 2 == 0) {
      total += ceil(len, 2) * x;
      total -= (len / 2) * x;
    } else {
      total -= ceil(len, 2) * x;
      total += (len / 2) * x;
    }
    cout << find() << '\n';
  }
  return 0;
}
