#include <bits/stdc++.h>
using namespace std;
const int maxn = 100100;
int A[maxn], B[maxn];
int main() {
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  for (int e = 0; e < n; e++) scanf("%d", A + e);
  for (int e = 0; e < m; e++) scanf("%d", B + e);
  long long sa = 0;
  for (int e = 0; e < n; e++)
    if (e & 1)
      sa -= A[e];
    else
      sa += A[e];
  set<long long> b;
  long long sb = 0;
  for (int e = 0; e < n; e++)
    if (e & 1)
      sb -= B[e];
    else
      sb += B[e];
  b.insert(sb);
  for (int e = 0;; e++) {
    if (e + n == m) break;
    sb -= B[e];
    sb *= -1;
    if (n & 1)
      sb += B[e + n];
    else
      sb -= B[e + n];
    b.insert(sb);
  }
  auto answer = [&](long long v) {
    auto it = b.lower_bound(v);
    if (it == b.end()) {
      it--;
      return abs(v - *it);
    }
    long long ans = abs(v - *it);
    if (it != b.begin()) {
      it--;
      ans = min(ans, abs(v - *it));
    }
    return ans;
  };
  printf("%lld\n", answer(sa));
  for (int e = 0; e < q; e++) {
    int l, r, a;
    scanf("%d %d %d", &l, &r, &a);
    if (l % 2 == r % 2) {
      if (l % 2)
        sa += a;
      else
        sa -= a;
    }
    printf("%lld\n", answer(sa));
  }
  return 0;
}

### Prompt

Generate a cpp solution to 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 maxn = 100100;
int A[maxn], B[maxn];
int main() {
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  for (int e = 0; e < n; e++) scanf("%d", A + e);
  for (int e = 0; e < m; e++) scanf("%d", B + e);
  long long sa = 0;
  for (int e = 0; e < n; e++)
    if (e & 1)
      sa -= A[e];
    else
      sa += A[e];
  set<long long> b;
  long long sb = 0;
  for (int e = 0; e < n; e++)
    if (e & 1)
      sb -= B[e];
    else
      sb += B[e];
  b.insert(sb);
  for (int e = 0;; e++) {
    if (e + n == m) break;
    sb -= B[e];
    sb *= -1;
    if (n & 1)
      sb += B[e + n];
    else
      sb -= B[e + n];
    b.insert(sb);
  }
  auto answer = [&](long long v) {
    auto it = b.lower_bound(v);
    if (it == b.end()) {
      it--;
      return abs(v - *it);
    }
    long long ans = abs(v - *it);
    if (it != b.begin()) {
      it--;
      ans = min(ans, abs(v - *it));
    }
    return ans;
  };
  printf("%lld\n", answer(sa));
  for (int e = 0; e < q; e++) {
    int l, r, a;
    scanf("%d %d %d", &l, &r, &a);
    if (l % 2 == r % 2) {
      if (l % 2)
        sa += a;
      else
        sa -= a;
    }
    printf("%lld\n", answer(sa));
  }
  return 0;
}

```