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[100009], b[100009], n, m, q;
int main() {
  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 sum = 0;
  for (int i = 0; i < n; i++) sum += (i & 1) ? -b[i] : b[i];
  s.insert(sum);
  for (int i = 1; i + n <= m; i++) {
    sum -= b[i - 1];
    sum = -sum;
    int end = i + n - 1;
    sum += (n & 1) ? b[end] : -b[end];
    s.insert(sum);
  }
  long long par = 0, impar = 0;
  for (int i = 0; i < n; i++)
    if (i & 1)
      impar += a[i];
    else
      par += a[i];
  q++;
  bool first = true;
  while (q--) {
    if (first)
      first = false;
    else {
      int a, b;
      long long x;
      scanf("%d %d %lld", &a, &b, &x);
      a--;
      b--;
      int len = b + 1 - a;
      int left = (len + 1) / 2;
      int right = len - left;
      if (a & 1) swap(left, right);
      par += left * x;
      impar += right * x;
    }
    long long total = par - impar;
    auto it = s.lower_bound(total);
    if (it == s.end()) {
      it--;
      printf("%lld\n", abs(total - *it));
    } else {
      auto it2 = it;
      if (it2 == s.begin())
        printf("%lld\n", abs(total - *it));
      else {
        it2--;
        long long op1 = abs(total - *it);
        long long op2 = abs(total - *it2);
        printf("%lld\n", min(op1, op2));
      }
    }
  }
}
