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;
template <typename T>
void _R(T &x) {
  cin >> x;
}
template <typename T>
void _R(vector<T> &x) {
  for (auto it = x.begin(); it != x.end(); it++) {
    _R(*it);
  }
}
void R() {}
template <typename T, typename... K>
void R(T &head, K &...tail) {
  _R(head);
  R(tail...);
}
template <typename T>
void _W(const T &x, const char c) {
  cout << x;
}
template <typename T>
void _W(const vector<T> &x, const char c) {
  for (auto it = x.cbegin(); it != x.cend(); it++) {
    if (it != x.cbegin()) putchar(c);
    _W(*it, c);
  }
}
void W() {}
template <typename T, typename... K>
void W(const T &head, const K &...tail) {
  _W(head, ' ');
  cout << (sizeof...(tail) ? ' ' : '\n') << flush;
  W(tail...);
}
const double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
template <typename T>
inline T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
inline void amin(T &x, T y) {
  if (x > y) x = y;
}
template <typename T>
inline void amax(T &x, T y) {
  if (x < y) x = y;
}
template <typename A, typename B>
class comp {
 public:
  bool operator()(const pair<A, B> &a, const pair<A, B> &b) {
    if (a.first != b.first)
      return a.first < b.first;
    else
      return a.first > b.first;
  }
};
int n, m, q;
vector<long long> a, b, sp;
vector<pair<long long, long long>> st;
void preprocess(void) { return; }
void build(int node, int l, int r) {
  if (l > r) {
    return;
  }
  if (l == r) {
    st[node] = {a[l], 0};
    return;
  }
  build(((node) << 1), l, (l + r) >> 1);
  build((((node) << 1) + 1), 1 + ((l + r) >> 1), r);
  st[node] = {st[((node) << 1)].first + st[(((node) << 1) + 1)].first, 0};
  return;
}
void update(int node, int l, int r, int i, int j, long long x) {
  if (st[node].second != 0) {
    if ((r - l + 1) % 2 == 0) {
      st[node].first += 0;
    } else {
      if ((l + 1) % 2 == 1) {
        st[node].first += st[node].second;
      } else {
        st[node].first -= st[node].second;
      }
    }
    if (l != r) {
      st[((node) << 1)].second += st[node].second;
      st[(((node) << 1) + 1)].second += st[node].second;
    }
    st[node].second = 0;
  }
  if (l > r || l > j || r < i) {
    return;
  }
  if (l >= i && r <= j) {
    if ((r - l + 1) % 2 == 0) {
      st[node].first += 0;
    } else {
      if ((l + 1) % 2 == 1) {
        st[node].first += x;
      } else {
        st[node].first -= x;
      }
    }
    if (l != r) {
      st[((node) << 1)].second += x;
      st[(((node) << 1) + 1)].second += x;
    }
    return;
  }
  update(((node) << 1), l, (l + r) >> 1, i, j, x);
  update((((node) << 1) + 1), 1 + ((l + r) >> 1), r, i, j, x);
  assert(st[((node) << 1)].second == 0);
  assert(st[(((node) << 1) + 1)].second == 0);
  st[node].first = st[((node) << 1)].first + st[(((node) << 1) + 1)].first;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  preprocess();
  int teeee;
  teeee = 1;
  for (int zeeee = (1); zeeee <= (teeee); zeeee++) {
    R(n, m, q);
    assert(m >= n);
    a.resize(n);
    b.resize(m + 1);
    R(a);
    long long suma = 0;
    for (int i = (0); i <= ((int)(n)-1); i++) {
      if ((i + 1) % 2 == 0) {
        a[i] = -a[i];
      }
      suma += a[i];
    }
    for (int i = (0); i <= ((int)(m)-1); i++) {
      R(b[i + 1]);
      if ((i + 1) % 2 == 0) {
        b[i + 1] = -b[i + 1];
      }
    }
    sp.resize(m - n + 1);
    vector<long long> psb(m + 1, 0);
    for (int i = (1); i <= (m); i++) {
      psb[i] = psb[i - 1] + b[i];
    }
    for (int j = (0); j <= (m - n); j++) {
      if (j % 2 == 0) {
        sp[j] = psb[n + j] - psb[j];
      } else {
        sp[j] = psb[j] - psb[n + j];
      }
    }
    sort(sp.begin(), sp.end());
    sp.erase(unique(sp.begin(), sp.end()), sp.end());
    st.resize(4 * n + 10);
    build(1, 0, n - 1);
    assert(st[1].first == suma && st[1].second == 0);
    ;
    for (int i = (0); i <= ((int)(q + 1) - 1); i++) {
      ;
      long long x;
      int l, r;
      if (i == 0) {
        l = 1;
        r = n;
        x = 0;
      } else {
        R(l, r, x);
      }
      l--, r--;
      if ((r - l + 1) % 2 == 1) {
        if ((l + 1) % 2 == 0) {
          suma -= x;
        } else {
          suma += x;
        }
      }
      update(1, 0, n - 1, l, r, x);
      long long s = st[1].first;
      assert(st[1].second == 0);
      if (s < 0) {
        s = -s;
        vector<long long>::iterator it = lower_bound(sp.begin(), sp.end(), -s);
        long long ans = LLONG_MAX;
        if (it == sp.end()) {
          it--;
        };
        if (it != sp.begin()) {
          for (vector<long long>::iterator it1 = it - 1;
               it1 < sp.end() && it1 <= it + 1; it1++) {
            amin(ans, llabs(s + (*it1)));
          }
        } else {
          for (vector<long long>::iterator it1 = it;
               it1 < sp.end() && it1 <= it + 1; it1++) {
            amin(ans, llabs(s + (*it1)));
          }
        }
        W(ans);
      } else {
        vector<long long>::iterator it = lower_bound(sp.begin(), sp.end(), s);
        long long ans = LLONG_MAX;
        if (it == sp.end()) {
          it--;
        };
        if (it != sp.begin()) {
          for (vector<long long>::iterator it1 = it - 1;
               it1 < sp.end() && it1 <= it + 1; it1++) {
            amin(ans, llabs(s - (*it1)));
          }
        } else {
          for (vector<long long>::iterator it1 = it;
               it1 < sp.end() && it1 <= it + 1; it1++) {
            amin(ans, llabs(s - (*it1)));
          }
        }
        W(ans);
      }
    }
  }
  cerr << (clock() / (double)CLOCKS_PER_SEC) << endl;
  return 0;
}
