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 long long MAXN = 2e5 + 100;
const long long maxn = 2e3 + 100;
const long long MAXn = 2e2 + 100;
const long long D = 1e9 + 7;
const long long INF = 1e18 + 7;
long long a[MAXN], b[MAXN];
long long k2[MAXN], k;
vector<long long> W;
int main() {
  long long n, m, q;
  cin >> n >> m >> q;
  long long k = 0, f = 1;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    k += f * a[i];
    f *= -1;
  }
  for (long long i = 0; i < m; i++) {
    cin >> b[i];
  }
  f = 1;
  for (long long i = 0; i < n; i++) {
    k2[0] += f * b[i];
    f *= -1;
  }
  f *= -1;
  for (long long i = n; i < m; i++) {
    k2[i - n + 1] = k2[i - n] * -1 + b[i - n] + (f * b[i]);
  }
  for (long long i = 0; i <= m - n; i++) {
    W.push_back(k2[i]);
  }
  sort(W.begin(), W.end());
  long long s = lower_bound(W.begin(), W.end(), k) - W.begin();
  if (s == W.size()) {
    cout << k - W[W.size() - 1] << endl;
  } else if (s > 0) {
    cout << min(abs(k - W[s]), abs(k - W[s - 1])) << endl;
  } else {
    cout << abs(k - W[s]) << endl;
  }
  long long x, l, r;
  for (long long i = 0; i < q; i++) {
    cin >> l >> r >> x;
    l--;
    if ((r - l) % 2 == 1) {
      if (l % 2) {
        k -= x;
      } else {
        k += x;
      }
    }
    s = lower_bound(W.begin(), W.end(), k) - W.begin();
    if (s == W.size()) {
      cout << k - W[W.size() - 1] << endl;
    } else if (s > 0) {
      cout << min(abs(k - W[s]), abs(k - W[s - 1])) << endl;
    } else {
      cout << abs(k - W[s]) << endl;
    }
  }
}
