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;
const int N = 1e5 + 228;
ll t;
ll sum[2][N];
vector<ll> b;
int n, m;
ll md(ll x) { return x < 0 ? -x : x; }
set<ll> s1, s2;
ll get_ans() {
  ll res = 2e18;
  if (!s1.empty()) {
    res = min({res, md(t - *s1.begin()), md(t - *s1.rbegin())});
  }
  if (!s2.empty()) {
    res = min({res, md(t + *s2.begin()), md(t + *s2.rbegin())});
  }
  if (t < 0) {
    auto it = s2.lower_bound(-t);
    if (it != s2.end()) res = min(res, md(t + *it));
    if (it != s2.begin()) res = min(res, md(t + *--it));
  } else {
    auto it = s1.lower_bound(t);
    if (it != s1.end()) res = min(res, md(t + -*it));
    if (it != s1.begin()) res = min(res, md(t - *--it));
  }
  return res;
}
int main() {
  int q;
  scanf("%d%d%d", &n, &m, &q);
  int qwe = 1;
  for (int i = 0; i < n; ++i) {
    ll x;
    scanf("%lld", &x);
    t += x * qwe;
    qwe = -qwe;
  }
  qwe = -1;
  for (int i = 0; i < m; ++i) {
    ll x;
    scanf("%lld", &x);
    sum[0][i] = x * qwe;
    qwe = -qwe;
    sum[1][i] = x * qwe;
    if (i > 0) {
      sum[0][i] += sum[0][i - 1];
      sum[1][i] += sum[1][i - 1];
    }
  }
  b.resize(m - n + 1);
  for (int i = n - 1, step = 0; step <= m - n; ++step, ++i) {
    b[step] = sum[step % 2][i];
    if (step > 0) b[step] -= sum[step % 2][step - 1];
    if (b[step] < 0)
      s1.insert(-b[step]);
    else
      s2.insert(b[step]);
  }
  sort(b.begin(), b.end());
  printf("%lld\n", get_ans());
  while (q--) {
    int l, r;
    ll x;
    scanf("%d%d", &l, &r);
    scanf("%lld", &x);
    int len = r - l + 1;
    if (len & 1) {
      if (l & 1)
        t += x;
      else
        t -= x;
    }
    printf("%lld\n", get_ans());
  }
}
