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 MAXN = 1e5 + 10;
long long v[MAXN];
set<long long> st;
long long sa;
long long solve() {
  long long ans = 1e16;
  auto it = st.lower_bound(-sa);
  if (it != st.end()) ans = min(ans, abs(sa + *it));
  if (it != st.begin()) {
    ans = min(ans, abs(sa + *prev(it)));
  }
  return ans;
}
int main() {
  int n, m, q;
  scanf(" %d %d", &(n), &(m));
  scanf(" %d", &(q));
  for (int i = 0; i < (n); ++i) {
    long long x;
    scanf(" %lld", &x);
    sa += x * (i & 1 ? -1 : 1);
  }
  long long sb = 0;
  for (int i = 0; i < (m); ++i) {
    long long x;
    scanf(" %lld", &x);
    v[i] = x;
    sb += x * (i & 1 ? -1 : 1);
    if (i >= n - 1) {
      int idx = i - n + 1;
      int sign = (idx & 1) ? 1 : -1;
      st.insert(sb * sign);
      sb -= (idx & 1 ? -1 : 1) * v[idx];
    }
  }
  printf("%lld\n", solve());
  for (int i = 0; i < (q); ++i) {
    int l, r;
    long long x;
    scanf(" %d %d", &(l), &(r));
    scanf(" %lld", &x);
    int sz = r - l + 1;
    long long p1 = ((sz / 2) + sz % 2) * x;
    long long p2 = ((sz / 2)) * x;
    sa += (l & 1 ? 1 : -1) * p1 + (l & 1 ? -1 : 1) * p2;
    printf("%lld\n", solve());
  }
  return 0;
}
