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;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int MAX = 1e5 + 10;
const long long mod = 1e9 + 7;
long long b[MAX], j[MAX];
set<long long> s;
set<long long>::iterator it;
int main() {
  long long n, m, q, i, a, x, temp, temp1, temp2, l, r, ans;
  while (~scanf("%lld%lld%lld", &n, &m, &q)) {
    a = 0;
    for (i = 1; i <= n; i++) {
      scanf("%lld", &x);
      if (i % 2)
        a += x;
      else
        a -= x;
    }
    temp = 0;
    b[0] = 0;
    for (i = 1; i <= m; i++) {
      scanf("%lld", &b[i]);
      if (i <= n) {
        if (i % 2)
          temp += b[i];
        else
          temp -= b[i];
        if (i == n) j[0] = temp;
      } else {
        temp -= b[i - n];
        temp *= -1;
        if (n % 2)
          temp += b[i];
        else
          temp -= b[i];
        j[i - n] = temp;
      }
    }
    for (i = 0; i <= m - n; i++) {
      s.insert(j[i]);
    }
    s.insert(LLINF);
    s.insert(-LLINF);
    it = s.upper_bound(a);
    it--;
    ans = min(abs(a - (*s.lower_bound(a))), abs(a - (*it)));
    printf("%lld\n", ans);
    while (q--) {
      scanf("%lld%lld%lld", &l, &r, &x);
      if ((r - l + 1) % 2) {
        if (l % 2)
          a += x;
        else
          a -= x;
      }
      it = s.upper_bound(a);
      it--;
      ans = min(abs(a - (*s.lower_bound(a))), abs(a - (*it)));
      printf("%lld\n", abs(ans));
    }
  }
  return 0;
}
