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 INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;
struct Rcompare {
  bool operator()(pair<int, int> lhs, pair<int, int> rhs) {
    return lhs.first > rhs.first;
  }
};
struct compare {
  bool operator()(pair<int, int> lhs, pair<int, int> rhs) {
    return lhs.first < rhs.first;
  }
};
std::vector<long long> ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, q;
  cin >> n >> m >> q;
  int a[n], b[m];
  long long sa = 0, sb = 0;
  bool fa = false, fb = false;
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    if (fa) {
      sa -= a[i];
    } else {
      sa += a[i];
    }
    fa = !fa;
  }
  for (int i = 0; i < (n); ++i) {
    cin >> b[i];
    if (!fb) {
      sb -= b[i];
    } else {
      sb += b[i];
    }
    fb = !fb;
  }
  ans.push_back(((sa + sb)));
  for (int i = 0; i < (m - n); ++i) {
    cin >> b[i + n];
    if (!fb) {
      sb -= b[i + n];
      if (n % 2 == 0) {
        sb += b[i];
      } else {
        sb -= b[i];
      }
    } else {
      sb += b[i + n];
      if (n % 2 != 0) {
        sb += b[i];
      } else {
        sb -= b[i];
      }
    }
    sb *= -1;
    ans.push_back(((sa + sb)));
  }
  sort(ans.begin(), ans.end());
  long long shf = 0;
  int ix = lower_bound(ans.begin(), ans.end(), shf) - ans.begin();
  if (ix != 0 && ix != m - n + 1) {
    cout << min(abs(ans[ix] - shf), abs(ans[ix - 1] - shf)) << "\n";
  } else if (ix == 0) {
    cout << abs(ans[ix] - shf) << "\n";
  } else {
    cout << abs(ans[ix - 1] - shf) << "\n";
  }
  for (int i = 0; i < (q); ++i) {
    int st, en, ch;
    cin >> st >> en >> ch;
    st = st % 2;
    en = en % 2;
    if (st == en) {
      if (st == 1) {
        shf -= ch;
      } else {
        shf += ch;
      }
    }
    int ix = lower_bound(ans.begin(), ans.end(), shf) - ans.begin();
    if (ix != 0 && ix != m - n + 1) {
      cout << min(abs(ans[ix] - shf), abs(ans[ix - 1] - shf)) << "\n";
    } else if (ix == 0) {
      cout << abs(ans[ix] - shf) << "\n";
    } else {
      cout << abs(ans[ix - 1] - shf) << "\n";
    }
  }
  return 0;
}
