#include <bits/stdc++.h>
using namespace std;
template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
const int N = 2e5 + 7;
int n, m, q;
long long a[N], b[N], F[N];
set<long long> S, R;
long long ro = 0;
long long solve() {
  auto it1 = S.lower_bound(ro);
  long long ans = 1e18 + 7;
  if (it1 != S.end()) ans = min(ans, abs(*it1 - ro));
  it1 = R.lower_bound(-ro);
  if (it1 != R.end()) ans = min(ans, abs(-ro - *it1));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  for (int i = 1; i <= n; i++)
    (i & 1) ? F[0] += a[i] - b[i] : F[0] += b[i] - a[i];
  long long upd = 0, zn = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (i == 1 || i == n + 1)
      upd += zn * b[i];
    else
      upd += zn * 2LL * b[i];
    zn *= (-1);
  }
  zn *= (-1);
  for (int j = 1; j <= m - n; j++) {
    F[j] = F[j - 1] + upd;
    upd -= b[j] - b[j + 1];
    if (zn == 1)
      upd += b[j + n] - b[j + n + 1];
    else
      upd += b[j + n + 1] - b[j + n];
    upd *= (-1);
  }
  for (int j = 0; j <= m - n; j++) {
    S.insert(F[j]);
    R.insert(-F[j]);
  }
  cout << solve() << "\n";
  for (int i = 0; i < q; i++) {
    int l, r;
    long long x;
    cin >> l >> r >> x;
    if (l % 2 == r % 2) {
      if (l % 2 == 1)
        ro -= x;
      else
        ro += x;
    }
    cout << solve() << "\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;
template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
const int N = 2e5 + 7;
int n, m, q;
long long a[N], b[N], F[N];
set<long long> S, R;
long long ro = 0;
long long solve() {
  auto it1 = S.lower_bound(ro);
  long long ans = 1e18 + 7;
  if (it1 != S.end()) ans = min(ans, abs(*it1 - ro));
  it1 = R.lower_bound(-ro);
  if (it1 != R.end()) ans = min(ans, abs(-ro - *it1));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  for (int i = 1; i <= n; i++)
    (i & 1) ? F[0] += a[i] - b[i] : F[0] += b[i] - a[i];
  long long upd = 0, zn = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (i == 1 || i == n + 1)
      upd += zn * b[i];
    else
      upd += zn * 2LL * b[i];
    zn *= (-1);
  }
  zn *= (-1);
  for (int j = 1; j <= m - n; j++) {
    F[j] = F[j - 1] + upd;
    upd -= b[j] - b[j + 1];
    if (zn == 1)
      upd += b[j + n] - b[j + n + 1];
    else
      upd += b[j + n + 1] - b[j + n];
    upd *= (-1);
  }
  for (int j = 0; j <= m - n; j++) {
    S.insert(F[j]);
    R.insert(-F[j]);
  }
  cout << solve() << "\n";
  for (int i = 0; i < q; i++) {
    int l, r;
    long long x;
    cin >> l >> r >> x;
    if (l % 2 == r % 2) {
      if (l % 2 == 1)
        ro -= x;
      else
        ro += x;
    }
    cout << solve() << "\n";
  }
  return 0;
}

```