#include <bits/stdc++.h>
using namespace std;
const long long lt = 100005;
long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m, q;
  cin >> n >> m >> q;
  vector<long long> a, b;
  long long i, val;
  for (i = 0; i < n; i++) {
    cin >> val;
    a.push_back(val);
  }
  for (i = 0; i < m; i++) {
    cin >> val;
    b.push_back(val);
  }
  vector<long long> pre;
  long long temp = 0;
  long long x = 0;
  for (i = 0; i < n; i++) {
    if (i % 2 == 1) {
      temp += b[i];
      x -= a[i];
    } else {
      temp -= b[i];
      x += a[i];
    }
  }
  pre.push_back(temp);
  for (i = 0; i < m - n; i++) {
    temp += b[i];
    temp = -temp;
    if (n % 2 == 0)
      temp += b[i + n];
    else
      temp -= b[i + n];
    pre.push_back(temp);
  }
  sort(pre.begin(), pre.end());
  long long w = -x;
  long long v = lower_bound(pre.begin(), pre.end(), w) - pre.begin();
  if (v == (long long)(pre).size()) v--;
  long long ans = min(abs(x + pre[v]), abs(x + pre[v - 1]));
  cout << ans << "\n";
  while (q--) {
    long long l, r, t;
    cin >> l >> r >> t;
    if (l % 2 == 0) x -= t;
    if (r % 2 == 1) x += t;
    w = -x;
    v = lower_bound(pre.begin(), pre.end(), w) - pre.begin();
    if (v == (long long)(pre).size()) v--;
    ans = min(abs(x + pre[v]), abs(x + pre[v - 1]));
    cout << ans << "\n";
  }
  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 long long lt = 100005;
long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m, q;
  cin >> n >> m >> q;
  vector<long long> a, b;
  long long i, val;
  for (i = 0; i < n; i++) {
    cin >> val;
    a.push_back(val);
  }
  for (i = 0; i < m; i++) {
    cin >> val;
    b.push_back(val);
  }
  vector<long long> pre;
  long long temp = 0;
  long long x = 0;
  for (i = 0; i < n; i++) {
    if (i % 2 == 1) {
      temp += b[i];
      x -= a[i];
    } else {
      temp -= b[i];
      x += a[i];
    }
  }
  pre.push_back(temp);
  for (i = 0; i < m - n; i++) {
    temp += b[i];
    temp = -temp;
    if (n % 2 == 0)
      temp += b[i + n];
    else
      temp -= b[i + n];
    pre.push_back(temp);
  }
  sort(pre.begin(), pre.end());
  long long w = -x;
  long long v = lower_bound(pre.begin(), pre.end(), w) - pre.begin();
  if (v == (long long)(pre).size()) v--;
  long long ans = min(abs(x + pre[v]), abs(x + pre[v - 1]));
  cout << ans << "\n";
  while (q--) {
    long long l, r, t;
    cin >> l >> r >> t;
    if (l % 2 == 0) x -= t;
    if (r % 2 == 1) x += t;
    w = -x;
    v = lower_bound(pre.begin(), pre.end(), w) - pre.begin();
    if (v == (long long)(pre).size()) v--;
    ans = min(abs(x + pre[v]), abs(x + pre[v - 1]));
    cout << ans << "\n";
  }
  return 0;
}

```