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>
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  int n, m, q;
  std::cin >> n >> m >> q;
  std::vector<int> a(n), b(m);
  for (int i = 0; i < n; ++i) std::cin >> a[i];
  for (int i = 0; i < m; ++i) std::cin >> b[i];
  int64_t A = 0;
  for (int i = 0; i < n; ++i) {
    if (i % 2 == 0)
      A += a[i];
    else
      A -= a[i];
  }
  std::vector<int64_t> B(m - n + 1, 0);
  for (int i = 0; i < n; ++i) {
    if (i % 2 == 0)
      B[0] -= b[i];
    else
      B[0] += b[i];
  }
  for (int j = 1; j <= m - n; ++j) {
    B[j] = -(B[j - 1] + b[j - 1]);
    if (n % 2 == 0)
      B[j] += b[j + n - 1];
    else
      B[j] -= b[j + n - 1];
  }
  std::vector<int64_t> pos, neg;
  for (auto u : B) {
    u += A;
    if (u > 0)
      pos.push_back(u);
    else
      neg.push_back(-u);
  }
  sort(pos.begin(), pos.end());
  sort(neg.begin(), neg.end());
  int64_t dif = 0;
  if (pos.size() && neg.size())
    std::cout << std::min(pos[0], neg[0]) << "\n";
  else if (pos.size())
    std::cout << pos[0] << "\n";
  else
    std::cout << neg[0] << "\n";
  while (q--) {
    int l, r, x;
    std::cin >> l >> r >> x;
    if ((r - l) % 2 == 0) {
      if (l % 2)
        dif += x;
      else
        dif -= x;
    }
    int64_t ans = 1e15;
    if (dif > 0) {
      if (pos.size()) ans = pos[0] + dif;
      if (neg.size()) {
        int64_t L;
        int v = lower_bound(neg.begin(), neg.end(), dif) - neg.begin();
        if (v == 0)
          L = neg[0] - dif;
        else if (v == (int)neg.size())
          L = dif - *neg.rbegin();
        else
          L = std::min(neg[v] - dif, dif - neg[v - 1]);
        ans = std::min(ans, L);
      }
    } else {
      if (neg.size()) ans = neg[0] - dif;
      if (pos.size()) {
        int64_t R;
        int v = lower_bound(pos.begin(), pos.end(), -dif) - pos.begin();
        if (v == 0)
          R = pos[0] + dif;
        else if (v == (int)pos.size())
          R = -dif - *pos.rbegin();
        else
          R = std::min(pos[v] + dif, -dif - pos[v - 1]);
        ans = std::min(ans, R);
      }
    }
    std::cout << ans << "\n";
  }
  return 0;
}
