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 n, m, q, a[100101], b[100100], i, j, l, r, v;
long long sa, sb[100100], minb, maxb;
long long cauta() {
  long long* p = lower_bound(sb + 1, sb + m - n + 2, sa);
  long long poz = p - sb;
  if (poz == 1)
    return sb[1] - sa;
  else if (poz == m - n + 2)
    return sa - sb[poz - 1];
  else
    return min(abs(sb[poz] - sa), abs(sa - sb[poz - 1]));
}
int main() {
  cin >> n >> m >> q;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= m; i++) cin >> b[i];
  long long sgn = 1;
  for (i = 1; i <= n; i++) {
    sa += a[i] * sgn;
    sgn = -sgn;
  }
  sgn = 1;
  for (i = 1; i <= n; i++) {
    sb[1] += sgn * b[i];
    sgn = -sgn;
  }
  sgn = -sgn;
  for (i = 2; i <= m - n + 1; i++) {
    sb[i] = -sb[i - 1] + b[i - 1] + sgn * b[i + n - 1];
  }
  sort(sb + 1, sb + m - n + 2);
  cout << cauta() << '\n';
  for (i = 1; i <= q; i++) {
    cin >> l >> r >> v;
    if ((r - l + 1) % 2 == 1) {
      if (l % 2 == 1)
        sa += v;
      else
        sa -= v;
    }
    cout << cauta() << '\n';
  }
}
