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;
long long calc(long long asum, long long bsum) { return abs(asum + bsum); }
long long getmin(long long asum, vector<long long> &jvals) {
  long long l = 1;
  long long r = (long long)jvals.size() - 1;
  long long ans = INT_MAX;
  while (l <= r) {
    long long mid = l + (r - l) / 2;
    long long left = calc(asum, jvals[mid - 1]);
    long long current = calc(asum, jvals[mid]);
    long long right = calc(asum, jvals[mid + 1]);
    if (left >= current && current <= right) {
      ans = current;
      break;
    } else if (left >= current && current >= right)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
signed main(void) {
  cin.tie(0)->sync_with_stdio(0);
  long long n, m, q;
  cin >> n >> m >> q;
  long long asum = 0;
  long long p = 1;
  for (long long i = 0; i < n; i++) {
    long long a;
    cin >> a;
    asum += a * p - a * (p ^ 1);
    p ^= 1;
  }
  vector<long long> b(m);
  vector<long long> jvals;
  set<long long> jset;
  for (auto &x : b) cin >> x;
  long long window = 0;
  for (long long i = 0; i < n; i++)
    window += b[i] * (i % 2 == 1) - b[i] * (i % 2 == 0);
  jset.insert(window);
  p = 0;
  for (long long i = n; i < m; i++) {
    window += b[i] * (i % 2 == 1) - b[i] * (i % 2 == 0);
    long long r = i - n;
    window -= b[r] * (r % 2 == 1) - b[r] * (r % 2 == 0);
    jset.insert(window * p - window * (p ^ 1));
    p ^= 1;
  }
  jset.insert(-1e18);
  jset.insert(+1e18);
  for (auto x : jset) jvals.push_back(x);
  cout << getmin(asum, jvals) << "\n";
  for (long long i = 0; i < q; i++) {
    long long l, r, x;
    cin >> l >> r >> x;
    if ((r - l) % 2 == 0) asum += x * (l % 2 == 1) - x * (l % 2 == 0);
    cout << getmin(asum, jvals) << "\n";
  }
}
