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 = 200200;
const int mod = 1e9 + 7;
int a[N], b[N], m, q;
long long t[4 * N];
int n;
set<long long> all;
long long suma = 0;
inline long long get_ans() {
  long long res = 1e18;
  long long f = -suma;
  auto it = all.lower_bound(f);
  if (it != all.end()) res = min(res, abs(suma + *it));
  if (it != all.begin()) {
    it--;
    res = min(res, abs(suma + *it));
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i % 2 == 1)
      suma += a[i];
    else
      suma -= a[i];
  }
  for (int i = 1; i <= m; ++i) cin >> b[i];
  long long best = 1e18;
  long long odd = 0, even = 0;
  for (int i = m; i >= 1; --i) {
    if (i % 2 == 1) {
      odd -= b[i];
      even += b[i];
    } else {
      odd += b[i];
      even -= b[i];
    }
    if (i + n <= m) {
      int w = i + n;
      if (w % 2 == 1) {
        odd += b[w];
        even -= b[w];
      } else {
        odd -= b[w];
        even += b[w];
      }
    }
    if (i + n - 1 <= m) {
      if (i % 2 == 1)
        all.insert(odd);
      else
        all.insert(even);
    }
  }
  long long res = get_ans();
  cout << res << "\n";
  while (q--) {
    int l, r, add;
    cin >> l >> r >> add;
    long long len = r - l + 1;
    if (l % 2 == 1) {
      long long p = (len + 1) / 2;
      long long m = len - p;
      suma += p * add;
      suma -= m * add;
    } else {
      long long p = (len) / 2;
      long long m = len - p;
      suma += p * add;
      suma -= m * add;
    }
    long long res = get_ans();
    cout << res << "\n";
  }
  return 0;
}
