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;
vector<long long> v;
vector<long long>::iterator it;
int a[100005], b[100005];
int main() {
  std::ios::sync_with_stdio(false);
  int n, m, i, x, q, fl, l, r;
  long long ans, iinf = (1000 * 1000 * 1000 + 5);
  iinf *= (1000 * 1000 * 1000 + 5);
  ans = iinf;
  long long val = 0, s = 0;
  cin >> n >> m >> q;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2) {
      val -= a[i];
    } else {
      val += a[i];
    }
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  for (i = 0; i < n; i++) {
    if (i % 2) {
      s -= b[i];
      fl = 0;
    } else {
      s += b[i];
      fl = 1;
    }
  }
  v.push_back(s);
  for (i = n; i < m; i++) {
    s -= b[i - n];
    if (fl == 1) {
      s -= b[i];
    } else {
      s += b[i];
    }
    s *= -1;
    v.push_back(s);
  }
  sort(v.begin(), v.end());
  it = lower_bound(v.begin(), v.end(), val);
  if (it != v.end()) {
    ans = min(ans, abs(val - *it));
  }
  if (it != v.begin()) {
    it--;
    ans = min(ans, abs(val - *it));
  }
  cout << ans << endl;
  while (q--) {
    cin >> l >> r >> x;
    l--;
    r--;
    ans = iinf;
    if (((r - l + 1) % 2) == 0) {
      it = lower_bound(v.begin(), v.end(), val);
      if (it != v.end()) {
        ans = min(ans, abs(val - *it));
      }
      if (it != v.begin()) {
        it--;
        ans = min(ans, abs(val - *it));
      }
      cout << ans << endl;
    } else {
      if (l % 2 == 0) {
        it = lower_bound(v.begin(), v.end(), val + x);
        if (it != v.end()) {
          ans = min(ans, abs(val - *it + x));
        }
        if (it != v.begin()) {
          it--;
          ans = min(ans, abs(val - *it + x));
        }
        cout << ans << endl;
        val += x;
      } else {
        it = lower_bound(v.begin(), v.end(), val - x);
        if (it != v.end()) {
          ans = min(ans, abs(val - *it - x));
        }
        if (it != v.begin()) {
          it--;
          ans = min(ans, abs(val - *it - x));
        }
        cout << ans << endl;
        val -= x;
      }
    }
  }
}
