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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int maxn = 1e5 + 10;
const int maxx = 600005;
const double EPS = 1e-8;
const double eps = 1e-8;
const int mod = 1e9 + 7;
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) {
    return 0;
  }
  while (c != '-' && (c < '0' || c > '9')) {
    c = getchar();
  }
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
  }
  ret *= sgn;
  return 1;
}
inline bool scan_lf(double &num) {
  char in;
  double Dec = 0.1;
  bool IsN = false, IsD = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && in != '.' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else if (in == '.') {
    IsD = true;
    num = 0;
  } else
    num = in - '0';
  if (!IsD) {
    while (in = getchar(), in >= '0' && in <= '9') {
      num *= 10;
      num += in - '0';
    }
  }
  if (in != '.') {
    if (IsN) num = -num;
    return true;
  } else {
    while (in = getchar(), in >= '0' && in <= '9') {
      num += Dec * (in - '0');
      Dec *= 0.1;
    }
  }
  if (IsN) num = -num;
  return true;
}
void Out(long long a) {
  if (a < 0) {
    putchar('-');
    a = -a;
  }
  if (a >= 10) Out(a / 10);
  putchar(a % 10 + '0');
}
void print(long long a) { Out(a), puts(""); }
long long sum[maxn], sum1[maxn], sum2[maxn];
long long a[maxn], b[maxn];
long long ans = 0;
int tot;
int n, m, q;
long long solve() {
  int l = 1, r = tot;
  int len = 100;
  long long val = LINF;
  while (l < r) {
    int mid = l + r + 1 >> 1;
    if (ans - sum[mid] >= 0)
      l = mid;
    else
      r = mid - 1;
  }
  val = min(val, abs(ans - sum[l]));
  if (l + 1 <= tot) val = min(val, abs(ans - sum[l + 1]));
  return val;
}
int main() {
  while (scanf("%d%d%d", &n, &m, &q) != EOF) {
    ans = 0;
    long long cur = 1;
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &a[i]);
      ans += cur * a[i];
      cur *= -1;
    }
    sum1[0] = 0;
    sum2[0] = 0;
    for (int i = 1; i <= m; i++) {
      scanf("%I64d", &b[i]);
      if (i & 1) {
        if (i == 1) {
          sum1[i] = b[i];
        } else
          sum1[i] = sum1[i - 2] + b[i];
        sum2[i] = sum2[i - 1];
      } else {
        sum2[i] = sum2[i - 2] + b[i];
        sum1[i] = sum1[i - 1];
      }
    }
    for (int i = 1; i <= m - n + 1; i++) {
      int r = i + n - 1;
      if (i & 1) {
        sum[i] = sum1[r] - sum1[i - 1] - (sum2[r] - sum2[i - 1]);
      } else {
        sum[i] = sum2[r] - sum2[i - 1] - (sum1[r] - sum1[i - 1]);
      }
    }
    tot = m - n + 1;
    sort(sum + 1, sum + 1 + tot);
    print(solve());
    while (q--) {
      int l, r;
      long long x;
      scanf("%d%d", &l, &r);
      scanf("%I64d", &x);
      if ((r - l + 1) & 1) {
        if (l & 1)
          ans += x;
        else
          ans -= x;
      }
      print(solve());
    }
  }
}
