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;
void pass() { return; }
const double EPS = 1e-6;
mt19937 rnd((rand() << 15 + rand()));
const long long MOD = 1e9 + 7;
const long long BIG_MOD = 1e18 + 7;
const long long MAXN = 1e5 + 7;
long long a[MAXN], b[MAXN], pref0[MAXN], pref1[MAXN];
vector<long long> v = {-BIG_MOD, BIG_MOD};
void print(long long sum) {
  long long l = 0;
  long long r = v.size() - 1;
  while (l < r - 1) {
    long long mid = (l + r) / 2;
    if (sum + v[mid] <= 0) {
      l = mid;
    } else {
      r = mid;
    }
  }
  cout << min(abs(v[l] + sum), abs(v[l + 1] + sum)) << '\n';
}
signed main() {
  cout.tie(NULL);
  cout.precision(20);
  ios_base::sync_with_stdio(false);
  long long n, m, q;
  cin >> n >> m >> q;
  long long sum = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    if ((i + 1) % 2) {
      sum += a[i];
    } else {
      sum -= a[i];
    }
  }
  for (long long i = 0; i < m; ++i) {
    cin >> b[i];
    if ((i + 1) % 2) {
      pref0[i + 1] = pref0[i] + b[i];
      pref1[i + 1] = pref1[i];
    } else {
      pref0[i + 1] = pref0[i];
      pref1[i + 1] = pref1[i] + b[i];
    }
    if (i >= n - 1) {
      long long l = i + 1 - n + 1;
      long long sum0 = pref0[i + 1] - pref0[l - 1];
      long long sum1 = pref1[i + 1] - pref1[l - 1];
      if (l % 2) {
        v.push_back(sum1 - sum0);
      } else {
        v.push_back(sum0 - sum1);
      }
    }
  }
  sort(v.begin(), v.end());
  print(sum);
  long long l, r, x;
  for (long long i = 0; i < q; ++i) {
    cin >> l >> r >> x;
    long long len = r - l + 1;
    if (l % 2) {
      sum += x * (len % 2);
    } else {
      sum -= x * (len % 2);
    }
    print(sum);
  }
  return 0;
}
