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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long N = 1e5 + 5;
long long g[N], a[N], b[N];
void solve() {
  long long n, m, q;
  cin >> n >> m >> q;
  long long s = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 0) {
      s += a[i];
    } else {
      s -= a[i];
    }
  }
  long long t = 0;
  for (long long i = 0; i < m; i++) {
    cin >> b[i];
    if (i >= n) continue;
    if (i % 2 == 0) {
      t += b[i];
    } else {
      t -= b[i];
    }
  }
  set<long long> f;
  for (long long i = 0; i < m - n + 1; i++) {
    g[i] = s - t;
    t -= b[i];
    t *= -1;
    if (n % 2) {
      t += b[i + n];
    } else {
      t -= b[i + n];
    }
    f.insert(g[i]);
  }
  long long tot = 0;
  for (long long i = 0; i < q + 1; i++) {
    long long ans = INF;
    auto it = f.lower_bound(-tot);
    if (it != f.end()) {
      amin(ans, *it + tot);
    }
    if (it != f.begin()) {
      it--;
      amin(ans, -tot - *it);
    }
    cout << ans << "\n";
    if (i < q) {
      long long l, r, x;
      cin >> l >> r >> x;
      l--;
      r--;
      if ((r - l + 1) % 2) {
        if (l % 2)
          tot -= x;
        else
          tot += x;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
