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.
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 10, INF = 1e18;
long long a[N], b[N];
long long f[N], X, n, m;
int tp;
inline void solve() {
  long long foo = -X, ans = INF;
  int id = lower_bound(f, f + m - n + 1, foo) - f;
  if (id <= m - n) ans = min(ans, f[id] + X);
  --id;
  if (id >= 0) ans = min(ans, -(f[id] + X));
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> n >> m >> q;
  if (n % 2 == 0)
    tp = -1;
  else
    tp = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 0)
      X += a[i];
    else
      X -= a[i];
  }
  for (int j = 0; j < m; j++) cin >> b[j];
  for (int j = m - n; j >= 0; j--) {
    if (j == m - n) {
      for (int k = j; k < j + n; k++) {
        if ((k - j) % 2 == 0)
          f[j] -= b[k];
        else
          f[j] += b[k];
      }
    } else {
      f[j] = -b[j] - f[j + 1] - 1LL * tp * b[j + n];
    }
  }
  sort(f, f + m - n + 1);
  solve();
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    --l;
    --r;
    if ((r - l) % 2 == 0) {
      if (l % 2 == 0)
        X += x;
      else
        X -= x;
    }
    solve();
  }
}
