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 maxn = 100005;
const int64_t inf = numeric_limits<int64_t>::min();
int n, m, q;
array<int64_t, maxn> a, b, even, odd, sumb;
int64_t query(int l, int r) {
  int64_t se = even[r], so = odd[r];
  if (l > 0) {
    se -= even[l - 1];
    so -= odd[l - 1];
  }
  if (l % 2 == 0) return se - so;
  return so - se;
}
int64_t abs1(int64_t a) { return a < 0 ? -a : a; }
int64_t query2(const set<pair<int64_t, int64_t> > &st, int64_t S) {
  int64_t ans = numeric_limits<int64_t>::max();
  auto it = st.lower_bound({S, inf});
  if (it != st.end()) ans = min(ans, abs1(S - it->first));
  if (it != st.begin()) {
    --it;
    ans = min(ans, abs1(S - it->first));
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> q;
  int64_t S = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i % 2 == 0)
      S += a[i];
    else
      S -= a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
    if (i > 0) {
      even[i] = even[i - 1];
      odd[i] = odd[i - 1];
    }
    if (i % 2 == 0) {
      even[i] += b[i];
    } else {
      odd[i] += b[i];
    }
  }
  set<pair<int64_t, int64_t> > st;
  for (int i = 0; i <= m - n; ++i) {
    sumb[i] = query(i, i + n - 1);
    if (i == 0)
      st.insert({sumb[i], inf});
    else
      st.insert({sumb[i], sumb[i - 1]});
  }
  cout << query2(st, S) << '\n';
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    l -= 1;
    r -= 1;
    if ((r - l) % 2 == 1) x = 0;
    if (l % 2 == 1) x = -x;
    S += x;
    cout << query2(st, S) << '\n';
  }
}
