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;
ofstream outfile;
ifstream infile;
vector<long long int> ans;
long long int a[100010], b[100010];
int n, m, q;
long long int find(vector<long long int>::iterator x) {
  if (x == ans.end()) return 1000000000000000000ll;
  return *x;
}
void solve(long long int search) {
  vector<long long int>::iterator iter =
      lower_bound(ans.begin(), ans.end(), -search);
  long long int val1 = 1000000000000000000ll, val2 = 1000000000000000000ll,
                val3 = 1000000000000000000ll;
  val1 = find(iter);
  if (iter != ans.end()) {
    ++iter;
    val2 = find(iter);
    --iter;
  }
  if (iter != ans.begin()) {
    --iter;
    val3 = *iter;
    ++iter;
  }
  cout << min({abs(val1 + search), abs(val2 + search), abs(val3 + search)})
       << "\n";
}
int main() {
  infile.open("input.txt");
  outfile.open("output.txt");
  ios::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  ;
  int i, j, l, r;
  long long int t, sum = 0, sum2 = 0;
  cin >> n >> m >> q;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i & 1)
      sum += a[i];
    else
      sum -= a[i];
  }
  for (i = 1; i <= m; ++i) {
    cin >> b[i];
  }
  for (i = 1; i <= n; ++i) {
    if (i & 1)
      sum2 -= b[i];
    else
      sum2 += b[i];
  }
  ans.push_back(sum2);
  for (j = 1; j <= m - n; ++j) {
    sum2 = -ans[j - 1];
    sum2 -= b[j];
    if (n & 1)
      sum2 -= b[j + n];
    else
      sum2 += b[j + n];
    ans.push_back(sum2);
  }
  for (i = 0; i <= m - n; ++i) ans[i] += sum;
  sort(ans.begin(), ans.end());
  sum = 0;
  solve(sum);
  for (i = 1; i <= q; ++i) {
    cin >> l >> r >> t;
    int siz = r - l + 1;
    if (!(siz % 2)) {
      solve(sum);
      continue;
    }
    if (l % 2) {
      sum += t;
    } else {
      sum -= t;
    }
    solve(sum);
  }
}
