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;
int64_t getmin(int64_t sum, vector<int64_t>& c) {
  int64_t p = lower_bound(c.begin(), c.end(), sum) - c.begin();
  return (min(sum - c[p - 1], c[p] - sum));
}
int main() {
  int64_t n, m, q;
  cin >> n >> m >> q;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; ++i) cin >> a[i];
  vector<int64_t> b(m);
  for (int64_t i = 0; i < m; ++i) cin >> b[i];
  vector<int64_t> c;
  int64_t val0 = 0;
  for (int64_t i = 0; i < n; ++i) {
    if (i % 2 == 0) {
      val0 += b[i];
    } else
      val0 -= b[i];
  }
  int64_t prev = (n) % 2;
  c.push_back(val0);
  for (int64_t i = 1; i < m - n + 1; ++i) {
    val0 = -val0;
    val0 += b[i - 1];
    if (prev)
      val0 += b[i + n - 1];
    else
      val0 -= b[i + n - 1];
    c.push_back(val0);
  }
  sort(c.begin(), c.end());
  c.push_back(1e17);
  vector<int64_t> d = {-100000000000000000};
  for (auto elem : c) {
    d.push_back(elem);
  }
  c = d;
  int64_t a1 = 0;
  for (int64_t i = 0; i < n; ++i) {
    if (i % 2 == 0)
      a1 += a[i];
    else
      a1 -= a[i];
  }
  cout << getmin(a1, c) << "\n";
  while (q--) {
    int64_t l, r, x;
    cin >> l >> r >> x;
    if (l & 1) a1 += x;
    if ((r + 1) & 1) a1 -= x;
    cout << getmin(a1, c) << endl;
  }
}
