#include <bits/stdc++.h>
using namespace std;
const int64_t N = 100 * 1000;
int64_t a[N + 3];
int64_t b[N + 3];
int64_t f(int64_t n) {
  if (n < 0) return -n;
  return n;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int64_t n, m, q;
  cin >> n >> m >> q;
  for (int64_t i = 1; i <= n; ++i) cin >> a[i];
  for (int64_t i = 1; i <= m; ++i) cin >> b[i];
  int64_t sum = 0;
  for (int64_t i = 1; i <= n; ++i) sum += pow(-1, i - 1) * a[i];
  vector<int64_t> B;
  int64_t val = 0;
  for (int64_t i = 1; i <= n; ++i) val += pow(-1, i) * b[i];
  B.push_back(val);
  for (int64_t i = n + 1; i <= m; ++i) {
    val -= pow(-1, i - n) * b[i - n];
    val += pow(-1, i) * b[i];
    if ((i - n + 1) % 2 == 0)
      B.push_back(-val);
    else
      B.push_back(val);
  }
  sort(B.begin(), B.end());
  int64_t ans = 1e18;
  for (auto it : B) {
    ans = min(ans, f(sum + it));
  }
  cout << ans << "\n";
  while (q--) {
    int64_t l, r, x;
    cin >> l >> r >> x;
    int64_t sz = r - l + 1;
    x = -x;
    if (sz & 1LL) {
      if (l & 1LL) {
        x = -x;
        sum += x;
      } else {
        sum += x;
      }
      ans = 1e18;
      int64_t pos = lower_bound(B.begin(), B.end(), -sum) - B.begin();
      if (pos == B.size()) pos--;
      ans = min(ans, f(sum + B[pos]));
      pos--;
      if (pos >= 0) ans = min(ans, f(sum + B[pos]));
      cout << ans << "\n";
    } else {
      cout << ans << "\n";
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int64_t N = 100 * 1000;
int64_t a[N + 3];
int64_t b[N + 3];
int64_t f(int64_t n) {
  if (n < 0) return -n;
  return n;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int64_t n, m, q;
  cin >> n >> m >> q;
  for (int64_t i = 1; i <= n; ++i) cin >> a[i];
  for (int64_t i = 1; i <= m; ++i) cin >> b[i];
  int64_t sum = 0;
  for (int64_t i = 1; i <= n; ++i) sum += pow(-1, i - 1) * a[i];
  vector<int64_t> B;
  int64_t val = 0;
  for (int64_t i = 1; i <= n; ++i) val += pow(-1, i) * b[i];
  B.push_back(val);
  for (int64_t i = n + 1; i <= m; ++i) {
    val -= pow(-1, i - n) * b[i - n];
    val += pow(-1, i) * b[i];
    if ((i - n + 1) % 2 == 0)
      B.push_back(-val);
    else
      B.push_back(val);
  }
  sort(B.begin(), B.end());
  int64_t ans = 1e18;
  for (auto it : B) {
    ans = min(ans, f(sum + it));
  }
  cout << ans << "\n";
  while (q--) {
    int64_t l, r, x;
    cin >> l >> r >> x;
    int64_t sz = r - l + 1;
    x = -x;
    if (sz & 1LL) {
      if (l & 1LL) {
        x = -x;
        sum += x;
      } else {
        sum += x;
      }
      ans = 1e18;
      int64_t pos = lower_bound(B.begin(), B.end(), -sum) - B.begin();
      if (pos == B.size()) pos--;
      ans = min(ans, f(sum + B[pos]));
      pos--;
      if (pos >= 0) ans = min(ans, f(sum + B[pos]));
      cout << ans << "\n";
    } else {
      cout << ans << "\n";
    }
  }
  return 0;
}

```