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;
using ll = long long int;
using pii = pair<int, int>;
using piii = pair<int, pii>;
using vi = vector<int>;
using vii = vector<pii>;
const int INF = 0x3F3F3F3F;
const ll INFL = 0x3F3F3F3F3F3F3F3FLL;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const int MAX = 1e5 + 2;
int N, M, Q;
ll arr[MAX], brr[MAX];
set<ll> first;
void calcF() {
  ll totA = accumulate(arr + 1, arr + N + 1, 0LL);
  ll totB = accumulate(brr + 1, brr + N + 1, 0LL);
  first.insert(totA + totB);
  for (int i = 1; i <= M - N; ++i) {
    totB += brr[N + i];
    totB -= brr[i];
    ll ans = totB;
    if (i % 2 == 1) {
      ans *= -1;
    }
    first.insert(totA + ans);
  }
}
ll solve(ll baseline) {
  auto it = first.upper_bound(baseline);
  ll ub = (it != first.end() ? *it : INFL) - baseline;
  ll lb = (it != first.begin() ? *(--it) : -INFL) - baseline;
  return min(abs(ub), abs(lb));
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> Q;
  for (int i = 1; i <= N; ++i) {
    cin >> arr[i];
    if (i % 2 == 0) {
      arr[i] *= -1;
    }
  }
  for (int i = 1; i <= M; ++i) {
    cin >> brr[i];
    if (i % 2 == 1) {
      brr[i] *= -1;
    }
  }
  calcF();
  cout << solve(0) << '\n';
  int l, r, x;
  ll baseline = 0;
  for (int i = 1; i <= Q; ++i) {
    cin >> l >> r >> x;
    int len = r - l + 1;
    if (len % 2 != 0) {
      if (l % 2 == 1) {
        baseline -= x;
      } else {
        baseline += x;
      }
    }
    cout << solve(baseline) << '\n';
  }
}
