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 int N = 100001;
int n, m, q, a[N], b[N];
long long mn, mx;
long long dp[2][N + 3];
vector<long long> all;
long long calc(long long s) {
  auto it = lower_bound(all.begin(), all.end(), -s);
  if (it != all.begin()) --it;
  long long res = 1e18;
  for (int i = 0; i < 2 && it != all.end(); ++i) {
    res = min(res, abs(s + *it));
    ++it;
  }
  return res;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  long long s = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    if (i % 2 == 1)
      s -= a[i];
    else
      s += a[i];
  }
  for (int i = 0; i < m; ++i) scanf("%d", b + i);
  long long mn = 1e18, mx = -1e18;
  for (int i = m - 1; i >= 0; --i) {
    dp[0][i] = dp[1][i + 1] + b[i];
    dp[1][i] = dp[0][i + 1] - b[i];
    if (i + n <= m) {
      long long cur = dp[1][i];
      if (n % 2 == 0)
        cur += dp[0][i + n];
      else
        cur -= dp[0][i + n];
      all.push_back(cur);
      mn = min(mn, cur);
      mx = max(mx, cur);
    }
  }
  sort(all.begin(), all.end());
  int l, r, val;
  {
    long long A = s + mn;
    long long B = s + mx;
    if (A < 0) A = -A;
    if (B < 0) B = -B;
    printf("%lld\n", calc(s));
  }
  while (q--) {
    scanf("%d %d %d", &l, &r, &val);
    int len = r - l + 1;
    long long neg = len / 2;
    long long pos = len - neg;
    if (l % 2 == 0) swap(neg, pos);
    s -= val * (long long)neg;
    s += val * (long long)pos;
    long long A = s + mn;
    long long B = s + mx;
    if (A < 0) A = -A;
    if (B < 0) B = -B;
    printf("%lld\n", calc(s));
  }
  return 0;
}
