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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using pi = pair<int, int>;
using vpi = vector<pair<int, int>>;
template <typename T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = 1e9 + 7;
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  vi a(n), b(m);
  for (int &e : a) cin >> e;
  for (int &e : b) cin >> e;
  set<ll> c;
  ll sum = 0;
  for (int i = 0; i < n; i++) sum += (i & 1 ? -b[i] : b[i]);
  c.insert(sum);
  for (int i = n; i < m; i++)
    sum = -sum + b[i - n] + (n & 1 ? b[i] : -b[i]), c.insert(sum);
  ;
  sum = 0;
  for (int i = 0; i < n; i++) sum += (i & 1 ? -a[i] : a[i]);
  ;
  auto it = c.lower_bound(sum);
  ll res = LINF;
  if (it != c.end()) {
    res = min(res, llabs(*it - sum));
  }
  if (it != c.begin()) {
    it--;
    res = min(res, llabs(*it - sum));
  }
  cout << res << "\n";
  while (q--) {
    ll l, r, x;
    cin >> l >> r >> x;
    if (l % 2 == r % 2) {
      sum += (l & 1 ? x : -x);
    }
    auto it = c.lower_bound(sum);
    ll res = LINF;
    if (it != c.end()) {
      res = min(res, llabs(*it - sum));
    }
    if (it != c.begin()) {
      it--;
      res = min(res, llabs(*it - sum));
    }
    cout << res << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int testcase = 1;
  while (testcase--) {
    solve();
  }
}
