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 = 1e5 + 5;
int n, m, q;
long long sum_a = 0, sum_b[N];
long long ar[N];
long long calc(int l) {
  int r = l + n - 1;
  long long ans = sum_b[r];
  if (l) ans -= sum_b[l - 1];
  if (l % 2) ans = -ans;
  return ans;
}
void output(void) {
  int x = upper_bound(ar, ar + m - n + 1, -sum_a) - ar;
  if (x == m - n + 1)
    cout << abs(sum_a + ar[x - 1]) << "\n";
  else if (x == 0)
    cout << abs(sum_a + ar[x]) << "\n";
  else
    cout << min(abs(sum_a + ar[x - 1]), abs(sum_a + ar[x])) << "\n";
}
void update(int l, int x, int v) {
  if (v % 2) {
    if (l % 2)
      sum_a -= x;
    else
      sum_a += x;
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> q;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    if (i % 2) a = -a;
    sum_a += a;
  }
  for (int i = 0; i < m; ++i) {
    int b;
    cin >> b;
    if (i % 2 == 0) b = -b;
    sum_b[i] = (i ? sum_b[i - 1] : 0) + b;
  }
  for (int i = 0; i < m - n + 1; ++i) {
    ar[i] = calc(i);
  }
  sort(ar, ar + m - n + 1);
  output();
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    --l, --r;
    update(l, x, (r - l + 1));
    output();
  }
  return 0;
}
