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;
signed main() {
  long long n, m, q;
  cin >> n >> m >> q;
  vector<long long> A(n);
  for (long long i = 0; i < n; ++i) {
    cin >> A[i];
  }
  vector<long long> B(m);
  for (long long i = 0; i < m; ++i) {
    cin >> B[i];
  }
  vector<long long> J(m - n + 1);
  long long cur = 0;
  for (long long i = 0; i < n; ++i) {
    if (i % 2 == 0) {
      cur += B[i];
    } else {
      cur -= B[i];
    }
  }
  J[0] = cur;
  for (long long i = 1; i < J.size(); ++i) {
    J[i] = B[i - 1] - J[i - 1];
    if (n % 2 == 0) {
      J[i] -= B[i + n - 1];
    } else {
      J[i] += B[i + n - 1];
    }
  }
  long long base = 0;
  for (long long i = 0; i < n; ++i) {
    if (i % 2 == 0) {
      base += A[i];
    } else {
      base -= A[i];
    }
  }
  for (long long i = 0; i < J.size(); ++i) {
    J[i] = base - J[i];
  }
  sort(J.begin(), J.end());
  long long qu = 0;
  {
    auto iter = upper_bound(J.begin(), J.end(), qu);
    long long mini = 1e18;
    if (iter != J.end()) {
      mini = min(mini, abs(*iter));
    }
    --iter;
    mini = min(mini, abs(*iter));
    cout << mini << endl;
  }
  for (long long i = 0; i < q; ++i) {
    long long l, r, x;
    cin >> l >> r >> x;
    --l;
    if ((r - l) % 2) {
      if (l % 2 == 0) {
        qu -= x;
      } else {
        qu += x;
      }
    }
    auto iter = upper_bound(J.begin(), J.end(), qu);
    long long mini = 1e18;
    if (iter != J.end()) {
      mini = min(mini, abs(*iter - qu));
    }
    --iter;
    mini = min(mini, abs(*iter - qu));
    cout << mini << endl;
  }
  return 0;
}
