#include <bits/stdc++.h>
using namespace std;
int a[1000 * 1000], b[1000 * 1000], n, m, q;
set<long long> r;
long long get(long long x) {
  set<long long>::iterator it = r.upper_bound(x);
  long long ans = abs(x - *it);
  if (it != r.begin()) {
    it--;
    ans = min(ans, x - *it);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  long long s = 0, p = 1, g1 = 0, g2 = 0;
  r.insert(1e17);
  r.insert(-1e17);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    s += a[i] * p;
    p *= -1;
  }
  p = 1;
  for (int j = 1; j <= m; j++) {
    cin >> b[j];
    g1 += b[j] * p;
    g2 += b[j] * p * (-1);
    if (j >= n) {
      if ((j - n) % 2 == 0) {
        r.insert(g1);
        g1 -= b[j - n + 1];
        g2 += b[j - n + 1];
      } else {
        r.insert(g2);
        g1 += b[j - n + 1];
        g2 -= b[j - n + 1];
      }
    }
    p *= -1;
  }
  cout << get(s) << '\n';
  long long l, r, x;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r >> x;
    if ((r - l + 1) % 2 != 0) {
      if (l & 1) {
        s += x;
      } else {
        s -= x;
      }
    }
    cout << get(s) << '\n';
  }
}

### 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;
int a[1000 * 1000], b[1000 * 1000], n, m, q;
set<long long> r;
long long get(long long x) {
  set<long long>::iterator it = r.upper_bound(x);
  long long ans = abs(x - *it);
  if (it != r.begin()) {
    it--;
    ans = min(ans, x - *it);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  long long s = 0, p = 1, g1 = 0, g2 = 0;
  r.insert(1e17);
  r.insert(-1e17);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    s += a[i] * p;
    p *= -1;
  }
  p = 1;
  for (int j = 1; j <= m; j++) {
    cin >> b[j];
    g1 += b[j] * p;
    g2 += b[j] * p * (-1);
    if (j >= n) {
      if ((j - n) % 2 == 0) {
        r.insert(g1);
        g1 -= b[j - n + 1];
        g2 += b[j - n + 1];
      } else {
        r.insert(g2);
        g1 += b[j - n + 1];
        g2 -= b[j - n + 1];
      }
    }
    p *= -1;
  }
  cout << get(s) << '\n';
  long long l, r, x;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r >> x;
    if ((r - l + 1) % 2 != 0) {
      if (l & 1) {
        s += x;
      } else {
        s -= x;
      }
    }
    cout << get(s) << '\n';
  }
}

```