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 long long INF = 2e18;
const int INFi = 2e9;
const int maxN = 2501;
const int P = 998244353;
const int md = 1e9 + 7;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> a(n), b(m);
  for (int i = 0; (i) < n; ++(i)) cin >> a[i];
  for (int i = 0; (i) < m; ++(i)) cin >> b[i];
  vector<long long> f(m - n + 1);
  long long s1 = 0;
  for (int i = 0; (i) < n; ++(i)) {
    if (i & 1)
      s1 -= a[i];
    else
      s1 += a[i];
  }
  long long s2 = 0;
  for (int i = 0; (i) < n; ++(i)) {
    if (i & 1)
      s2 += b[i];
    else
      s2 -= b[i];
  }
  f[0] = s1 + s2;
  for (int i = n; i < m; ++i) {
    s2 += b[i - n];
    s2 = -s2;
    if (n & 1) {
      s2 -= b[i];
    } else {
      s2 += b[i];
    }
    f[i - n + 1] = s1 + s2;
  }
  sort((f).begin(), (f).end());
  long long w = 0;
  auto get = [&w, &f]() {
    int lb = lower_bound((f).begin(), (f).end(), w) - f.begin();
    long long res = INF;
    if (lb != f.size()) {
      res = f[lb] - w;
    }
    if (lb != 0) {
      res = min(res, w - f[lb - 1]);
    }
    return res;
  };
  cout << get() << "\n";
  for (int _ = 0; (_) < q; ++(_)) {
    int l, r, x;
    cin >> l >> r >> x;
    if ((r - l + 1) & 1) {
      if (l & 1)
        w -= x;
      else
        w += x;
    }
    cout << get() << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int tests = 1;
  for (int _ = 0; (_) < tests; ++(_)) {
    solve();
  }
  return 0;
}
