#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 22;
int n, m;
vector<long long> vec;
long long a[N], b[N];
long long ask(long long x) {
  int id = lower_bound(vec.begin(), vec.end(), x) - vec.begin(), id2 = id - 1;
  if (id2 == -1) return abs(vec[id] - x);
  if (id == vec.size()) return abs(vec[id2] - x);
  return min(abs(vec[id] - x), abs(vec[id2] - x));
}
int main() {
  ios::sync_with_stdio(false);
  long long x = 0, y = 0, q;
  cin >> n >> m >> q;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (i % 2)
      x -= a[i];
    else
      x += a[i];
  }
  for (int i = 0; i < (m); i++) {
    cin >> b[i];
    if (i % 2 && i < n)
      y -= b[i];
    else if (i % 2 == 0 && i < n)
      y += b[i];
  }
  long long sw = 1;
  for (int i = 0; i <= m - n; i++) {
    vec.push_back(x - y * sw);
    if (i == m - n) break;
    if (i % 2)
      y += b[i];
    else
      y -= b[i];
    if ((i + n) % 2)
      y -= b[i + n];
    else
      y += b[i + n];
    sw *= -1;
  }
  sort(vec.begin(), vec.end());
  cout << ask(0) << endl;
  x = 0;
  while (q--) {
    long long l, r, val;
    cin >> l >> r >> val;
    l--, r--;
    if ((r - l + 1) % 2 == 1) {
      if (l % 2)
        x += val;
      else
        x -= val;
    }
    cout << ask(x) << endl;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
using namespace std;
const int N = 2e5 + 22;
int n, m;
vector<long long> vec;
long long a[N], b[N];
long long ask(long long x) {
  int id = lower_bound(vec.begin(), vec.end(), x) - vec.begin(), id2 = id - 1;
  if (id2 == -1) return abs(vec[id] - x);
  if (id == vec.size()) return abs(vec[id2] - x);
  return min(abs(vec[id] - x), abs(vec[id2] - x));
}
int main() {
  ios::sync_with_stdio(false);
  long long x = 0, y = 0, q;
  cin >> n >> m >> q;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (i % 2)
      x -= a[i];
    else
      x += a[i];
  }
  for (int i = 0; i < (m); i++) {
    cin >> b[i];
    if (i % 2 && i < n)
      y -= b[i];
    else if (i % 2 == 0 && i < n)
      y += b[i];
  }
  long long sw = 1;
  for (int i = 0; i <= m - n; i++) {
    vec.push_back(x - y * sw);
    if (i == m - n) break;
    if (i % 2)
      y += b[i];
    else
      y -= b[i];
    if ((i + n) % 2)
      y -= b[i + n];
    else
      y += b[i + n];
    sw *= -1;
  }
  sort(vec.begin(), vec.end());
  cout << ask(0) << endl;
  x = 0;
  while (q--) {
    long long l, r, val;
    cin >> l >> r >> val;
    l--, r--;
    if ((r - l + 1) % 2 == 1) {
      if (l % 2)
        x += val;
      else
        x -= val;
    }
    cout << ask(x) << endl;
  }
  return 0;
}

```