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 int N = 100005;
int a[N], b[N];
long long c[2], d[N];
set<long long> se;
long long mabs(long long x) { return x < 0 ? -x : x; }
long long calc(long long x) {
  long long res = 1ll << 60;
  auto it = se.upper_bound(x);
  if (it != se.end()) res = min(res, mabs(x - *it));
  --it;
  res = min(res, mabs(x - *it));
  return res;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= (n); ++i) cin >> a[i];
  for (int i = 1; i <= (m); ++i) cin >> b[i];
  for (int i = 1; i <= (n); ++i) c[i & 1] += a[i];
  for (int i = 1; i <= (m); ++i) d[i] = d[i - 1] + (i & 1 ? b[i] : -b[i]);
  for (int i = 1; i <= (m); ++i)
    if (i >= n) {
      if ((i - n) & 1)
        se.insert(d[i - n] - d[i]);
      else
        se.insert(d[i] - d[i - n]);
    }
  cout << calc(c[1] - c[0]) << endl;
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    if ((r - l + 1) & 1) c[l & 1] += x;
    cout << calc(c[1] - c[0]) << endl;
  }
  return 0;
}
