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;
class Solution {
 public:
  void solve(std::istream& in, std::ostream& out) {
    long long n, m, q;
    in >> n >> m >> q;
    std::vector<long long> a(n), b(m);
    std::copy_n(std::istream_iterator<long long>(in), n, a.begin());
    std::copy_n(std::istream_iterator<long long>(in), m, b.begin());
    long long f = 0;
    long long a_sum = 0;
    for (long long i = 0; i < n; ++i) a_sum += (i % 2 == 0 ? 1LL : -1LL) * a[i];
    for (long long i = 0; i < n; ++i) {
      f += (i % 2 == 0 ? -1LL : 1LL) * b[i];
    }
    std::vector<long long> vals(m - n + 1);
    vals[0] = a_sum + f;
    long long best_f = abs(a_sum + f);
    for (long long i = 0; i < m - n; ++i) {
      f = -b[i] - f + b[i + n] * (n % 2 == 0 ? 1LL : -1LL);
      vals[i + 1] = a_sum + f;
      best_f = std::min(abs(a_sum + f), best_f);
    }
    out << best_f << '\n';
    long long shift = 0;
    std::sort(vals.begin(), vals.end());
    for (long long i = 0; i < q; ++i) {
      long long l, r, x;
      in >> l >> r >> x;
      --l;
      --r;
      if (l % 2 == r % 2) {
        if (l % 2 == 0)
          shift -= x;
        else
          shift += x;
      }
      auto it = upper_bound(vals.begin(), vals.end(), shift);
      auto ans = std::numeric_limits<long long>::max();
      if (it != vals.end()) ans = std::min(ans, abs(*it - shift));
      if (it != vals.begin()) ans = std::min(ans, abs(*prev(it) - shift));
      out << ans << '\n';
    }
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}
