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 n, m, Q;
long long a[100005], b[100005];
long long asum[2][100005], bsum[2][100005];
vector<long long> vec;
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> Q;
  for (long long(i) = (1); (i) <= (n); (i)++) cin >> a[i];
  for (long long(i) = (1); (i) <= (m); (i)++) cin >> b[i];
  for (long long(i) = (1); (i) <= (n); (i)++) {
    asum[0][i] = asum[0][i - 1];
    asum[1][i] = asum[1][i - 1];
    asum[i % 2][i] += a[i];
  }
  for (long long(i) = (1); (i) <= (m); (i)++) {
    bsum[0][i] = bsum[0][i - 1];
    bsum[1][i] = bsum[1][i - 1];
    bsum[i % 2][i] += b[i];
  }
  long long sum = asum[1][n] - asum[0][n];
  for (long long(i) = (1); (i) <= (m); (i)++) {
    long long r = i + n - 1;
    if (r > m) break;
    vec.push_back(-(bsum[(i + 1) % 2][r] - bsum[(i + 1) % 2][i - 1]) +
                  (bsum[i % 2][r] - bsum[i % 2][i - 1]));
  }
  sort((vec).begin(), (vec).end());
  long long p = lower_bound((vec).begin(), (vec).end(), sum) - vec.begin();
  long long ans = 1e18;
  if (p < vec.size()) (ans) = min((ans), (abs(vec[p] - sum)));
  if (p - 1 >= 0) (ans) = min((ans), (abs(vec[p - 1] - sum)));
  cout << ans << "\n";
  long long l, r, x;
  for (long long(q) = (1); (q) <= (Q); (q)++) {
    cin >> l >> r >> x;
    long long e = r / 2 - (l - 1) / 2, o = r - l + 1 - e;
    ;
    sum += o * x - e * x;
    long long p = lower_bound((vec).begin(), (vec).end(), sum) - vec.begin();
    long long ans = 1e18;
    if (p < vec.size()) (ans) = min((ans), (abs(vec[p] - sum)));
    if (p - 1 >= 0) (ans) = min((ans), (abs(vec[p - 1] - sum)));
    cout << ans << "\n";
  }
  flush(cout);
  return 0;
}
