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, m, q, ta, tb, tc;
long long aa, bb[100005];
set<long long> s;
set<long long>::iterator ia, ib;
void go() {
  ia = s.lower_bound(aa);
  ib = ia;
  ia--;
  printf("%lld\n", min(aa - *ia, *ib - aa));
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ta);
    if (i % 2)
      aa -= ta;
    else
      aa += ta;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &ta);
    if (i % 2)
      bb[i + 1] = bb[i] - ta;
    else
      bb[i + 1] = bb[i] + ta;
  }
  for (int i = 0; i <= m - n; i++)
    if (i % 2)
      s.insert(-bb[i + n] + bb[i]);
    else
      s.insert(bb[i + n] - bb[i]);
  s.insert(LLONG_MAX / 3);
  s.insert(LLONG_MIN / 3);
  go();
  while (q--) {
    scanf("%d%d%d", &ta, &tb, &tc);
    if (ta % 2 == tb % 2)
      if (ta % 2)
        aa += tc;
      else
        aa -= tc;
    go();
  }
  return 0;
}
