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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long n) {
  a %= 1000000007;
  if (n == 1) return a;
  if (n == 0) return 1;
  if (n % 2)
    return (a * (power((a * a) % 1000000007, n / 2) % 1000000007)) % 1000000007;
  return power((a * a) % 1000000007, n / 2) % 1000000007;
}
const long long inf = (long long)1e18;
long long inverse(long long x) { return power(x, 1000000007 - 2) % 1000000007; }
void solve() {
  long long n, m, q, sum = 0;
  cin >> n >> m >> q;
  vector<long long> s(m + 5), a(n + 5), b(m + 5), neg, pos;
  for (long long i = 1; i < n + 1; ++i) {
    cin >> a[i];
    if (i % 2)
      sum += a[i];
    else
      sum -= a[i];
  }
  for (long long i = 1; i < m + 1; ++i) cin >> b[i];
  for (long long i = m; i > 0; --i)
    if (i % 2)
      s[i] = s[i + 1] + b[i];
    else
      s[i] = s[i + 1] - b[i];
  for (long long j = 0; j < m - n + 1; ++j) {
    long long x = s[j + 1] - s[j + n + 1];
    if (j % 2 == 0) x *= -1;
    if (x < 0)
      neg.push_back(-x);
    else if (x > 0)
      pos.push_back(x);
    else
      neg.push_back(x), pos.push_back(x);
  }
  sort(neg.begin(), neg.end());
  sort(pos.begin(), pos.end());
  long long l, r, x, ans = inf;
  if (sum >= 0) {
    if ((long long)pos.size()) ans = sum + pos[0];
    if ((long long)neg.size()) {
      long long j = lower_bound(neg.begin(), neg.end(), sum) - neg.begin();
      if (j < (long long)neg.size()) ans = min(ans, neg[j] - sum);
      if (j > 0) ans = min(ans, sum - neg[j - 1]);
    }
  } else {
    long long csum = -1 * sum;
    if ((long long)neg.size()) ans = csum + neg[0];
    if ((long long)pos.size()) {
      long long j = lower_bound(pos.begin(), pos.end(), csum) - pos.begin();
      if (j < (long long)pos.size()) ans = min(ans, pos[j] - csum);
      if (j > 0) ans = min(ans, csum - pos[j - 1]);
    }
  }
  cout << ans << "\n";
  while (q--) {
    ans = inf;
    cin >> l >> r >> x;
    if ((r - l) % 2 == 0) {
      if (l % 2)
        sum += x;
      else
        sum -= x;
    }
    if (sum >= 0) {
      if ((long long)pos.size()) ans = min(ans, sum + pos[0]);
      if ((long long)neg.size()) {
        long long j = lower_bound(neg.begin(), neg.end(), sum) - neg.begin();
        if (j < (long long)neg.size()) ans = min(ans, neg[j] - sum);
        if (j > 0) ans = min(ans, sum - neg[j - 1]);
      }
    }
    if (sum <= 0) {
      long long csum = -1 * sum;
      if ((long long)neg.size()) ans = min(ans, csum + neg[0]);
      if ((long long)pos.size()) {
        long long j = lower_bound(pos.begin(), pos.end(), csum) - pos.begin();
        if (j < (long long)pos.size()) ans = min(ans, pos[j] - csum);
        if (j > 0) ans = min(ans, csum - pos[j - 1]);
      }
    }
    cout << ans << "\n";
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  for (long long t = 1; t < T + 1; ++t) {
    solve();
  }
  return 0;
}
