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;
using uint = unsigned int;
using ll = long long;
using pii = pair<int, int>;
int a[100010], b[100010];
ll c[100010];
vector<ll> v;
ll query(ll s) {
  auto it = upper_bound(v.begin(), v.end(), s);
  auto prevIt = it;
  --prevIt;
  if (it == v.end()) return abs(s - (*prevIt));
  if (it == v.begin()) return abs(s - (*it));
  return min(abs(s - (*prevIt)), abs(s - (*it)));
}
int main() {
  int i, j, n, m, q, l, r, x;
  ll sa;
  cin >> n >> m >> q;
  for (i = 1; i <= n; ++i) cin >> a[i];
  for (j = 1; j <= m; ++j) cin >> b[j];
  for (sa = 0, i = 1; i <= n; ++i) sa += (i % 2 == 1 ? a[i] : -a[i]);
  for (j = 1; j <= n; ++j) c[1] += (j % 2 == 1 ? b[j] : -b[j]);
  for (i = 2; i <= m - n + 1; ++i)
    c[i] = -(c[i - 1] - b[i - 1]) + (n % 2 == 1 ? b[i + n - 1] : -b[i + n - 1]);
  for (i = 1; i <= m - n + 1; ++i) v.push_back(c[i]);
  sort(v.begin(), v.end());
  cout << query(sa) << '\n';
  while (q--) {
    cin >> l >> r >> x;
    if (l % 2 == r % 2) {
      if (l % 2 == 0)
        sa -= x;
      else
        sa += x;
    }
    cout << query(sa) << '\n';
  }
  return 0;
}
