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 main() {
  int T;
  T = 1;
  for (int t = 0; t != T; ++t) {
    int n, m, q;
    cin >> n >> m >> q;
    long long int a = 0;
    long long int sn = 1;
    for (int i = 0; i != n; ++i) {
      long long int x;
      cin >> x;
      a += sn * x;
      sn *= -1;
    }
    vector<long long int> b, bm;
    sn = 1;
    long long int sum = 0;
    for (int i = 0; i != m; ++i) {
      long long int x;
      cin >> x;
      b.push_back(x);
      if (i < n) {
        sum += sn * x;
        sn *= -1;
      } else {
        bm.push_back(sum);
        sum -= b[i - n];
        if (n % 2 == 0)
          sum += x;
        else
          sum -= x;
        sum *= -1;
      }
    }
    bm.push_back(sum);
    sort(bm.begin(), bm.end());
    auto it = lower_bound(bm.begin(), bm.end(), a);
    long long int ans;
    if (it == bm.end())
      ans = abs(*(it - 1) - a);
    else if (it == bm.begin())
      ans = abs(*(it)-a);
    else
      ans = min(abs(*(it - 1) - a), abs(*(it)-a));
    cout << ans << endl;
    for (int i = 0; i != q; ++i) {
      long long int l, r, x;
      cin >> l >> r >> x;
      if ((l - r) % 2 == 0) {
        if (l % 2 != 0)
          a += x;
        else
          a -= x;
      }
      auto it = lower_bound(bm.begin(), bm.end(), a);
      long long int ans;
      if (it == bm.end())
        ans = abs(*(it - 1) - a);
      else if (it == bm.begin())
        ans = abs(*(it)-a);
      else
        ans = min(abs(*(it - 1) - a), abs(*(it)-a));
      cout << ans << endl;
    }
  }
  return 0;
}
