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;
long long lazy[1000001];
long long tree[1000001];
set<long long> s;
long long a[1000001];
long long b[1000001];
void build(int x, int l, int r) {
  if (l == r) {
    if (l % 2)
      tree[x] = a[l];
    else
      tree[x] = -a[l];
  } else {
    build(x * 2, l, (l + r) / 2);
    build(x * 2 + 1, (l + r) / 2 + 1, r);
    tree[x] = tree[x * 2] + tree[x * 2 + 1];
  }
}
void lzy(int x, int l, int r) {
  if ((r - l + 1) % 2) {
    if (l % 2)
      tree[x] += lazy[x];
    else
      tree[x] -= lazy[x];
  }
  lazy[x * 2] += lazy[x];
  lazy[x * 2 + 1] += lazy[x];
  lazy[x] = 0;
}
void upd(int x, int l, int r, int s, int e, long long add) {
  lzy(x, l, r);
  if (l > e || s > r) return;
  if (l >= s && r <= e) {
    lazy[x] += add;
    lzy(x, l, r);
    return;
  }
  upd(x * 2, l, (l + r) / 2, s, e, add);
  upd(x * 2 + 1, (l + r) / 2 + 1, r, s, e, add);
  tree[x] = tree[x * 2] + tree[x * 2 + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
    if (i % 2 == 0) b[i] = -b[i];
    b[i] = b[i] + b[i - 1];
  }
  for (int i = n; i <= m; i++) {
    long long sum = b[i] - b[i - n];
    int l = i - n + 1;
    if (l % 2 == 0) sum = -sum;
    s.insert(sum);
  }
  build(1, 1, n);
  {
    long long sum = tree[1];
    long long best = 1e18;
    set<long long>::iterator it = s.upper_bound(sum);
    if (it != s.end()) {
      best = min(best, abs(sum - *it));
    }
    if (it != s.begin()) {
      it--;
      best = min(best, abs(sum - *it));
    }
    cout << best << endl;
  }
  while (q--) {
    int l, r, add;
    cin >> l >> r >> add;
    upd(1, 1, n, l, r, add);
    lzy(1, 1, n);
    long long sum = tree[1];
    long long best = 1e18;
    set<long long>::iterator it = s.upper_bound(sum);
    if (it != s.end()) {
      best = min(best, abs(sum - *it));
    }
    if (it != s.begin()) {
      it--;
      best = min(best, abs(sum - *it));
    }
    cout << best << endl;
  }
}
