#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, m, q, t, k = 0;
  cin >> n >> m >> q;
  t = m - n + 1;
  long long int a[n], b[m], c[t];
  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 % 2 == 0)
      k += a[i];
    else
      k -= a[i];
  }
  c[0] = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0)
      c[0] += b[i];
    else
      c[0] -= b[i];
  }
  for (int i = 1; i < t; i++) {
    c[i] = -c[i - 1] + b[i - 1];
    if (n % 2 == 0)
      c[i] -= b[n + i - 1];
    else
      c[i] += b[n + i - 1];
  }
  sort(c, c + t);
  if (k >= c[t - 1])
    cout << k - c[t - 1] << endl;
  else if (k <= c[0])
    cout << c[0] - k << endl;
  else {
    int p = 0, e = t, x;
    while (e - p > 1) {
      x = (e + p) / 2;
      if (k >= c[x])
        p = x;
      else
        e = x;
    }
    if (c[e] - k > k - c[p])
      cout << k - c[p] << endl;
    else
      cout << c[e] - k << endl;
  }
  for (int i = 0; i < q; i++) {
    int l, r, y;
    cin >> l >> r >> y;
    if (r % 2 == 0 && l % 2 == 0) k -= y;
    if (r % 2 != 0 && l % 2 != 0) k += y;
    if (k >= c[t - 1])
      cout << k - c[t - 1] << endl;
    else if (k <= c[0])
      cout << c[0] - k << endl;
    else {
      int p = 0, e = t, x;
      while (e - p > 1) {
        x = (e + p) / 2;
        if (k >= c[x])
          p = x;
        else
          e = x;
      }
      if (c[e] - k > k - c[p])
        cout << k - c[p] << endl;
      else
        cout << c[e] - k << endl;
    }
  }
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, m, q, t, k = 0;
  cin >> n >> m >> q;
  t = m - n + 1;
  long long int a[n], b[m], c[t];
  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 % 2 == 0)
      k += a[i];
    else
      k -= a[i];
  }
  c[0] = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0)
      c[0] += b[i];
    else
      c[0] -= b[i];
  }
  for (int i = 1; i < t; i++) {
    c[i] = -c[i - 1] + b[i - 1];
    if (n % 2 == 0)
      c[i] -= b[n + i - 1];
    else
      c[i] += b[n + i - 1];
  }
  sort(c, c + t);
  if (k >= c[t - 1])
    cout << k - c[t - 1] << endl;
  else if (k <= c[0])
    cout << c[0] - k << endl;
  else {
    int p = 0, e = t, x;
    while (e - p > 1) {
      x = (e + p) / 2;
      if (k >= c[x])
        p = x;
      else
        e = x;
    }
    if (c[e] - k > k - c[p])
      cout << k - c[p] << endl;
    else
      cout << c[e] - k << endl;
  }
  for (int i = 0; i < q; i++) {
    int l, r, y;
    cin >> l >> r >> y;
    if (r % 2 == 0 && l % 2 == 0) k -= y;
    if (r % 2 != 0 && l % 2 != 0) k += y;
    if (k >= c[t - 1])
      cout << k - c[t - 1] << endl;
    else if (k <= c[0])
      cout << c[0] - k << endl;
    else {
      int p = 0, e = t, x;
      while (e - p > 1) {
        x = (e + p) / 2;
        if (k >= c[x])
          p = x;
        else
          e = x;
      }
      if (c[e] - k > k - c[p])
        cout << k - c[p] << endl;
      else
        cout << c[e] - k << endl;
    }
  }
}

```