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 hsh[1000005] = {1}, hsh2[1000001] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1, i, j;
  long long temp = 524288;
  while (t--) {
    long long n, m, q;
    cin >> n >> m >> q;
    long long arr[n];
    long long sum = 0;
    for (i = 0; i < n; i++) {
      cin >> arr[i];
      if (i % 2 == 0)
        sum += arr[i];
      else
        sum -= arr[i];
    }
    long long brr[m], pre[m];
    for (i = 0; i < m; i++) {
      cin >> brr[i];
      if (i == 0)
        pre[i] = -brr[i];
      else {
        if (i % 2 == 0)
          pre[i] = pre[i - 1] - brr[i];
        else
          pre[i] = pre[i - 1] + brr[i];
      }
    }
    long long pre2[m - n + 1];
    set<long long> s;
    set<long long>::iterator it, it2;
    for (i = 0; i < (m - n + 1); i++) {
      pre2[i] = pre[n - 1 + i];
      if (i > 0) {
        if (i % 2 != 0) {
          pre2[i] = (-pre[n - 1 + i]);
          if (i > 0) pre2[i] = pre2[i] + pre[i - 1];
        } else {
          pre2[i] = (pre[n - 1 + i]);
          if (i > 0) pre2[i] = pre2[i] - pre[i - 1];
        }
      }
      s.insert(pre2[i]);
    }
    j = -sum;
    it = s.lower_bound(j);
    if (it == s.end()) it--;
    long long ans = LLONG_MAX;
    ans = min(ans, abs(sum + (*it)));
    it2 = it;
    it2++;
    if (it2 != s.end()) ans = min(ans, abs(sum + (*it2)));
    if (it != s.begin()) it--;
    ans = min(ans, abs(sum + (*it)));
    cout << ans << "\n";
    for (i = 0; i < q; i++) {
      long long l, r, k;
      cin >> l >> r >> k;
      long long len = r - l + 1;
      len %= 2;
      if (l % 2 == 0) {
        sum -= (k * len);
      } else {
        sum += (k * len);
      }
      j = -sum;
      it = s.lower_bound(j);
      if (it == s.end()) it--;
      ans = LLONG_MAX;
      ans = min(ans, abs(sum + (*it)));
      it2 = it;
      it2++;
      if (it2 != s.end()) ans = min(ans, abs(sum + (*it2)));
      if (it != s.begin()) it--;
      ans = min(ans, abs(sum + (*it)));
      cout << ans << "\n";
    }
  }
}
