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;
using ld = long double;
const ll mod = 1000000009;
const ll mxr = 1e6 + 10;
const ll oo = 33720360000000000;
const ll si = 200005;
vector<ll> pos;
ll sz;
void bsearch(ll x) {
  ll l = 0, r = sz - 1, ans = 1000000000000000000;
  while (l <= r) {
    ll mid = l + (r - l) / 2;
    if (pos[mid] + x > 0) {
      r = mid - 1;
      ans = min(ans, pos[mid] + x);
    } else if (pos[mid] + x < 0) {
      l = mid + 1;
      ans = min(ans, -(pos[mid] + x));
    } else {
      ans = 0;
      break;
    }
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n, m, q;
  cin >> n >> m >> q;
  ll b[m], pr[m], tot = 0;
  for (int i = 0; i < n; i++) {
    ll x;
    cin >> x;
    i % 2 ? tot -= x : tot += x;
  }
  for (int i = 0; i < m; i++) cin >> b[i];
  pr[0] = -b[0];
  for (int i = 1; i < m; i++) {
    pr[i] = pr[i - 1];
    i % 2 ? pr[i] += b[i] : pr[i] -= b[i];
  }
  pos.push_back(pr[n - 1]);
  for (int i = n; i < m; i++) {
    ll con = pr[i] - pr[i - n];
    (i - n + 1) % 2 ? pos.push_back(-con) : pos.push_back(con);
  }
  sort(pos.begin(), pos.end());
  sz = pos.size();
  bsearch(tot);
  while (q--) {
    ll l, r, x, us;
    cin >> l >> r >> x;
    l -= 1;
    r -= 1;
    us = r - l + 1;
    if (us % 2) {
      if (l % 2)
        tot -= x;
      else
        tot += x;
    }
    bsearch(tot);
  }
  return 0;
}
