#include <bits/stdc++.h>
const int md = 1e9 + 7;
const long long hs = 149;
using namespace std;
long long val1, sum = 0;
int n, m, q, l, r, x;
vector<int> a, b;
set<long long> val2;
long long solve() {
  auto it = val2.lower_bound(-val1);
  if (it == val2.end()) {
    auto temp = it;
    temp--;
    return abs(val1 + *temp);
  }
  if (it == val2.begin()) {
    if (val2.size() == 1) return abs(val1 + *it);
    auto temp = it;
    temp++;
    return min(abs(val1 + *it), abs(val1 + *temp));
  }
  auto itl = it, itr = it;
  itl--, itr++;
  if (itr == val2.end()) return min(abs(val1 + *it), abs(val1 + *itl));
  return min(min(abs(val1 + *itl), abs(val1 + *it)), abs(val1 + *itr));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m >> q;
  a.resize(n), b.resize(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    if (i & 1)
      sum -= a[i];
    else
      sum += a[i];
  }
  val1 = sum;
  sum = 0;
  for (int i = 0; i < n; i++) {
    if (i & 1)
      sum += b[i];
    else
      sum -= b[i];
  }
  val2.insert(sum);
  for (int i = 1; i <= m - n; i++) {
    sum += b[i - 1];
    sum *= -1;
    if (n & 1)
      sum -= b[i + n - 1];
    else
      sum += b[i + n - 1];
    val2.insert(sum);
  }
  cout << solve() << '\n';
  while (q--) {
    cin >> l >> r >> x;
    if ((l & 1) && (r & 1)) val1 += x;
    if (!(l & 1) && !(r & 1)) val1 -= x;
    cout << solve() << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
const int md = 1e9 + 7;
const long long hs = 149;
using namespace std;
long long val1, sum = 0;
int n, m, q, l, r, x;
vector<int> a, b;
set<long long> val2;
long long solve() {
  auto it = val2.lower_bound(-val1);
  if (it == val2.end()) {
    auto temp = it;
    temp--;
    return abs(val1 + *temp);
  }
  if (it == val2.begin()) {
    if (val2.size() == 1) return abs(val1 + *it);
    auto temp = it;
    temp++;
    return min(abs(val1 + *it), abs(val1 + *temp));
  }
  auto itl = it, itr = it;
  itl--, itr++;
  if (itr == val2.end()) return min(abs(val1 + *it), abs(val1 + *itl));
  return min(min(abs(val1 + *itl), abs(val1 + *it)), abs(val1 + *itr));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m >> q;
  a.resize(n), b.resize(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  for (int i = 0; i < n; i++) {
    if (i & 1)
      sum -= a[i];
    else
      sum += a[i];
  }
  val1 = sum;
  sum = 0;
  for (int i = 0; i < n; i++) {
    if (i & 1)
      sum += b[i];
    else
      sum -= b[i];
  }
  val2.insert(sum);
  for (int i = 1; i <= m - n; i++) {
    sum += b[i - 1];
    sum *= -1;
    if (n & 1)
      sum -= b[i + n - 1];
    else
      sum += b[i + n - 1];
    val2.insert(sum);
  }
  cout << solve() << '\n';
  while (q--) {
    cin >> l >> r >> x;
    if ((l & 1) && (r & 1)) val1 += x;
    if (!(l & 1) && !(r & 1)) val1 -= x;
    cout << solve() << '\n';
  }
  return 0;
}

```