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;
const int N = 1e+5 + 10;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  int64_t cur = 0;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    if (i & 1) a *= -1;
    cur += a;
  }
  vector<int> b(m);
  set<int64_t> fans;
  int64_t fcur = 0;
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
    if (i & 1) b[i] *= -1;
    fcur += b[i];
    if (i >= n) fcur -= b[i - n];
    if (i >= n - 1) {
      if ((i - n + 1) & 1)
        fans.insert(-fcur);
      else
        fans.insert(fcur);
    }
  }
  vector<int> l(q + 1), r(q + 1), x(q + 1);
  for (int i = 1; i <= q; ++i) {
    cin >> l[i] >> r[i] >> x[i];
  }
  for (int i = 0; i <= q; ++i) {
    --l[i];
    if ((r[i] - l[i]) & 1) {
      if (l[i] & 1) x[i] *= -1;
      cur += x[i];
    }
    auto it = fans.lower_bound(cur);
    int64_t ans = 1e+18;
    if (it != fans.end()) {
      ans = abs(cur - *it);
    }
    if (it != fans.begin()) {
      --it;
      ans = min(ans, abs(cur - *it));
    }
    cout << ans << '\n';
  }
  return 0;
}
