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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 200031;
int n, m, q, a[N], b[N];
long long S[N][2];
long long total_sum;
vector<long long> V;
int count_odd(int x) { return x / 2 + x % 2; }
int count_even(int x) { return x / 2; }
long long make_query(long long val) {
  long long res = 1e18;
  int id = lower_bound(V.begin(), V.end(), val) - V.begin();
  if (id != V.size()) res = min(res, V[id] - val);
  if (id != 0) {
    --id;
    res = min(res, val - V[id]);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  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++) {
    if (i % 2 == 1)
      total_sum += a[i];
    else
      total_sum -= a[i];
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= 1; j++) {
      S[i][j] = S[i - 1][j];
    }
    S[i][i % 2] += b[i];
  }
  for (int i = 0; i <= m - n; i++) {
    long long cur_sum = 0;
    cur_sum += S[i + n][1 - i % 2];
    cur_sum -= S[i][1 - i % 2];
    cur_sum -= S[i + n][i % 2];
    cur_sum += S[i][i % 2];
    V.push_back(cur_sum);
  }
  sort(V.begin(), V.end());
  cout << make_query(total_sum) << "\n";
  for (int i = 1; i <= q; i++) {
    int l, r, val;
    cin >> l >> r >> val;
    int cnt_odd = count_odd(r) - count_odd(l - 1);
    int cnt_even = count_even(r) - count_even(l - 1);
    total_sum += cnt_odd * 1ll * val;
    total_sum -= cnt_even * 1ll * val;
    cout << make_query(total_sum) << "\n";
  }
  cin.get();
  cin.get();
  return 0;
}
