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;
long long A, m;
long long b[100100];
long long B[100100];
void _find() {
  int in = 1, sf = m;
  while (in <= sf) {
    int mij = (in + sf) / 2;
    if (-A == B[mij]) {
      cout << 0 << '\n';
      return;
    }
    if (B[mij] < -A)
      in = mij + 1;
    else
      sf = mij - 1;
  }
  if (sf == 0)
    cout << abs(A + B[1]) << '\n';
  else if (in == m + 1)
    cout << abs(A + B[m]) << '\n';
  else
    cout << min(abs(A + B[in]), abs(A + B[sf])) << '\n';
}
int main() {
  int n, lst = -1, q, a, l, v, r;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    lst *= -1;
    A += lst * a;
  }
  lst = 1;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    lst *= -1;
    B[1] += lst * b[i];
  }
  for (int i = n + 1; i <= m; i++) {
    cin >> b[i];
    B[i - n + 1] = -B[i - n] - b[i - n] + lst * b[i];
  }
  m -= n;
  m++;
  sort(B + 1, B + m + 1);
  _find();
  for (int i = 1; i <= q; i++) {
    cin >> l >> r >> v;
    if ((r - l + 1) % 2 == 1) {
      if (r % 2)
        A += v;
      else
        A -= v;
    }
    _find();
  }
  return 0;
}
