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 "03"
const long long int MOD = 1e9 + 7;
const long long int INF = 1011111111;
const long long int LLINF = 1000111000111000111LL;
const long double EPS = 1e-10;
const long double PI = 3.14159265358979323;
using namespace std;
long long int power(long long int base, long long int exponent) {
  long long int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base);
    base = (1LL * base * base);
    exponent >>= 1;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k, q;
  cin >> n >> m >> q;
  long long int a[n + 1];
  long long int b[m + 1];
  vector<long long int> pre;
  long long int fa = 0;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= m; i++) cin >> b[i];
  long long int sum = 0;
  for (i = 1; i <= n; i++) {
    if (i & 1ll)
      fa += a[i];
    else
      fa -= a[i];
  }
  for (i = 1; i <= n; i++) {
    if (i & 1ll)
      sum -= b[i];
    else
      sum += b[i];
  }
  pre.push_back(sum);
  for (i = n + 1; i <= m; i++) {
    long long int x = sum + b[i - n];
    x *= -1ll;
    if (n & 1ll)
      x -= b[i];
    else
      x += b[i];
    pre.push_back(x);
    sum = x;
  }
  sort(pre.begin(), pre.end());
  long long int sz = pre.size();
  long long int to = -fa;
  long long int id = lower_bound(pre.begin(), pre.end(), to) - pre.begin();
  long long int ans = LLINF;
  for (i = max(0ll, id - 5); i < min(sz, id + 5); i++)
    ans = min(ans, abs(fa + pre[i]));
  cout << ans << '\n';
  while (q--) {
    long long int l, r, x;
    cin >> l >> r >> x;
    long long int cnt = r - l + 1;
    if (cnt % 2ll == 0)
      ;
    else {
      if (l % 2ll == 1)
        fa += x;
      else
        fa -= x;
    }
    long long int to = -fa;
    long long int id = lower_bound(pre.begin(), pre.end(), to) - pre.begin();
    long long int ans = LLINF;
    for (i = max(0ll, id - 5); i < min(sz, id + 5); i++)
      ans = min(ans, abs(fa + pre[i]));
    cout << ans << '\n';
  }
  cerr << '\n'
       << "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
}
