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 maxn = 1e5 + 100;
long long a[maxn], b[maxn];
long long zh[maxn];
vector<long long> v;
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (i % 2 != 0)
      sum += a[i];
    else
      sum -= a[i];
  }
  for (int i = 1; i <= m; i++) scanf("%lld", &b[i]);
  int num = 1;
  for (int i = m - n + 1; i <= m; i++) {
    if (num % 2 != 0)
      zh[m - n] = zh[m - n] - b[i];
    else
      zh[m - n] = zh[m - n] + b[i];
    num++;
  }
  v.push_back(zh[m - n]);
  for (int i = m - n - 1; i >= 0; i--) {
    if (n % 2 != 0)
      zh[i + 1] += b[i + n + 1];
    else
      zh[i + 1] -= b[i + n + 1];
    zh[i] = -zh[i + 1] - b[i + 1], v.push_back(zh[i]);
  }
  sort(v.begin(), v.end());
  int l = 1, r = 0, x;
  for (int i = 0; i <= q; i++) {
    if ((r - l) % 2 == 0 && l % 2 != 0)
      sum += x;
    else if ((r - l) % 2 == 0 && l % 2 == 0)
      sum -= x;
    int k = lower_bound(v.begin(), v.end(), -sum) - v.begin();
    long long ans = 0x3f3f3f3f3f3f3f3f;
    if (k == 0)
      ans = min(ans, abs(sum + v[k]));
    else if (k == v.size())
      ans = min(ans, abs(sum + v[k - 1]));
    else
      ans = min({ans, abs(sum + v[k - 1]), abs(sum + v[k])});
    cout << ans << endl;
    if (i != q) scanf("%d%d%d", &l, &r, &x);
  }
  return 0;
}
