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;
int n, v[100001], m, k, x;
long long p1, p2, b1, b2, l, r;
vector<long long> M;
long long ctb(long long x) {
  int st = 0, dr = M.size() - 1, mid;
  while (st <= dr) {
    mid = (st + dr) / 2;
    if (M[mid] > x)
      dr = mid - 1;
    else
      st = mid + 1;
  }
  if (dr < 0)
    return M[st] - x;
  else if (st >= M.size())
    return x - M[dr];
  return min(x - M[dr], M[st] - x);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (i % 2 == 1)
      p1 += x;
    else
      p2 += x;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    if (i % 2 == 1)
      b1 += v[i];
    else
      b2 += v[i];
  }
  for (int i = n + 1; i <= m; i++) {
    cin >> v[i];
    if ((i - n) % 2 == 0) {
      M.push_back(b1 - b2);
      b2 -= v[i - n];
    } else {
      M.push_back(b2 - b1);
      b1 -= v[i - n];
    }
    if (i % 2 == 1)
      b1 = b1 + v[i];
    else
      b2 = b2 + v[i];
  }
  if ((m + 1 - n) % 2 == 0)
    M.push_back(b1 - b2);
  else
    M.push_back(b2 - b1);
  sort(M.begin(), M.end());
  cout << ctb(p2 - p1) << '\n';
  for (int i = 1; i <= k; i++) {
    cin >> l >> r >> x;
    if (l % 2 == 1) {
      p1 += x;
      l++;
    }
    if (r % 2 == 0 && r >= l) {
      p2 += x;
      r--;
    }
    if (r > l) {
      p1 += (r - l + 1) / 2;
      p2 += (r - l + 1) / 2;
    }
    cout << ctb(p2 - p1) << '\n';
  }
  return 0;
}
