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 maxn = 1e5 + 10;
const int MOD = 1e9 + 7;
set<long long> bsums;
long long solve(long long sum_a) {
  auto idx = bsums.upper_bound(sum_a);
  if (idx == bsums.end()) return sum_a - *(--idx);
  if (idx == bsums.begin()) return *idx - sum_a;
  long long res = *idx - sum_a;
  --idx;
  return min(res, sum_a - *idx);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n, m, q;
  cin >> n >> m >> q;
  vector<long long> a(n), b(m);
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  long long sum_a = 0;
  vector<long long> sum_b(m - n + 1, 0);
  for (int i = 0; i < n; ++i) sum_a += (i % 2 == 0 ? 1 : -1) * a[i];
  for (int i = 0; i < n; ++i) sum_b[0] += (i % 2 == 0 ? 1 : -1) * b[i];
  for (int i = 1; i <= m - n; ++i) {
    sum_b[i] =
        -(sum_b[i - 1] - b[i - 1]) + (n % 2 == 0 ? -1 : 1) * b[i + n - 1];
  }
  for (int i = 0; i <= m - n; ++i) bsums.insert(sum_b[i]);
  cout << solve(sum_a) << "\n";
  for (int i = 0; i < q; ++i) {
    int l, r;
    long long x;
    cin >> l >> r >> x;
    if ((r - l + 1) & 1) {
      if (l & 1)
        sum_a += x;
      else
        sum_a -= x;
    }
    cout << solve(sum_a) << "\n";
  }
  return 0;
}
