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;
int a[100005];
int b[100005];
int main() {
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) scanf("%d", &b[i]);
  set<long long> s;
  long long now = 0;
  for (int i = 0; i < n; i++) {
    if (i & 1)
      now -= a[i];
    else
      now += a[i];
  }
  long long temp = 0;
  for (int i = 0; i < n - 1; i++) {
    if (i & 1)
      temp -= b[i];
    else
      temp += b[i];
  }
  for (int i = n - 1; i < m; i++) {
    if (n & 1)
      temp += b[i];
    else
      temp -= b[i];
    s.insert(temp);
    temp -= b[i - n + 1];
    temp = -temp;
  }
  for (int i = 0; i <= q; i++) {
    if (i != 0) {
      int l, r, first;
      scanf("%d %d %d", &l, &r, &first);
      if ((r - l + 1) % 2 == 1) {
        if (l & 1) {
          now += first;
        } else {
          now -= first;
        }
      }
    }
    set<long long>::iterator it = s.lower_bound(now);
    long long ans = 1e18;
    if (it != s.end()) {
      ans = min(*it - now, ans);
    }
    if (it != s.begin()) {
      it--;
      ans = min(now - *it, ans);
    }
    printf("%I64d\n", ans);
  }
}
