Problem: Problem

Given a sequence A with n elements, a sequence B with m elements, q queries each consisting of an integer c.
For each query
The absolute value of the difference between the sum of all [la, ra] in sequence A and the sum of all [lb, rb] in sequence B is c. La, ra, lb, rb (0 ≤ la ≤ ra ≤ ra ≤ Find the number of combinations of n−1, 0 ≤ lb ≤ rb ≤ m−1, sequence numbers start from 0).

Constraints

* 1 ≤ n, m ≤ 4 × 104
* 1 ≤ q ≤ 105
* 1 ≤ ai, bi ≤ 5
* 0 ≤ ci ≤ 2 × 105

Input


n m q
a0 a1 ... an−1
b0 b1 ... bm−1
c0
c1
...
cq−1


All inputs are given as integers.
The number of elements n and m in the sequence and the number of queries q are given in the first row.
The elements of sequence A are given in the second row, and the elements of sequence B are given in the third row, separated by blanks.
The value ci of each query is given from the 4th line to the q line.

Output

The output consists of q lines. Print the number of combinations for each query in turn on one line.

Examples

Input

3 3 1
1 2 3
3 1 2
3


Output

6


Input

5 4 2
1 2 3 4 5
2 2 2 2
11
12


Output

3
4

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;
using Int = long long;

namespace FFT
{
  using dbl = double;

  struct num
  {
    dbl x, y;

    num() { x = y = 0; }

    num(dbl x, dbl y) : x(x), y(y) {}
  };

  inline num operator+(num a, num b)
  {
    return num(a.x + b.x, a.y + b.y);
  }

  inline num operator-(num a, num b)
  {
    return num(a.x - b.x, a.y - b.y);
  }

  inline num operator*(num a, num b)
  {
    return num(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
  }

  inline num conj(num a)
  {
    return num(a.x, -a.y);
  }

  int base = 1;
  vector< num > rts = {{0, 0},
                       {1, 0}};
  vector< int > rev = {0, 1};

  const dbl PI = acosl(-1.0);

  void ensure_base(int nbase)
  {
    if(nbase <= base) return;

    rev.resize(1 << nbase);
    for(int i = 0; i < (1 << nbase); i++)
      rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));

    rts.resize(1 << nbase);
    while(base < nbase) {
      dbl angle = 2 * PI / (1 << (base + 1));
      for(int i = 1 << (base - 1); i < (1 << base); i++) {
        rts[i << 1] = rts[i];
        dbl angle_i = angle * (2 * i + 1 - (1 << base));
        rts[(i << 1) + 1] = num(cos(angle_i), sin(angle_i));
      }
      base++;
    }
  }

  void fft(vector< num > &a, int n = -1)
  {
    if(n == -1) n = a.size();
    assert((n & (n - 1)) == 0);

    int zeros = __builtin_ctz(n);
    ensure_base(zeros);
    int shift = base - zeros;
    for(int i = 0; i < n; i++)
      if(i < (rev[i] >> shift))
        swap(a[i], a[rev[i] >> shift]);

    for(int k = 1; k < n; k <<= 1) {
      for(int i = 0; i < n; i += 2 * k) {
        for(int j = 0; j < k; j++) {
          num z = a[i + j + k] * rts[j + k];
          a[i + j + k] = a[i + j] - z;
          a[i + j] = a[i + j] + z;
        }
      }
    }
  }

  vector< num > fa;

  vector< Int > multiply(vector< int > &a, vector< int > &b)
  {
    int need = a.size() + b.size() - 1;
    int nbase = 0;
    while((1 << nbase) < need) nbase++;
    ensure_base(nbase);

    int sz = 1 << nbase;
    if(sz > (int) fa.size()) fa.resize(sz);
    for(int i = 0; i < sz; i++) {
      int x = (i < (int) a.size() ? a[i] : 0);
      int y = (i < (int) b.size() ? b[i] : 0);
      fa[i] = num(x, y);
    }
    fft(fa, sz);

    num r(0, -0.25 / sz);
    for(int i = 0; i <= (sz >> 1); i++) {
      int j = (sz - i) & (sz - 1);
      num z = (fa[j] * fa[j] - conj(fa[i] * fa[i])) * r;
      if(i != j)
        fa[j] = (fa[i] * fa[i] - conj(fa[j] * fa[j])) * r;
      fa[i] = z;
    }
    fft(fa, sz);

    vector< Int > res(need);
    for(int i = 0; i < need; i++)
      res[i] = fa[i].x + 0.5;

    return res;
  }

};

int main()
{
  int N, M, Q, A[40001], B[40001], C[100000];
  vector< int > F(200001), G(200001);

  scanf("%d %d %d", &N, &M, &Q);
  for(int i = 0; i < N; i++) scanf("%d", &A[i + 1]), A[i + 1] += A[i];
  for(int i = 0; i < M; i++) scanf("%d", &B[i + 1]), B[i + 1] += B[i];
  for(int i = 0; i < Q; i++) scanf("%d", &C[i]);

  for(int i = 1; i <= N; i++) {
    for(int j = 0; j < i; j++) {
      ++F[A[i] - A[j]];
    }
  }
  for(int i = 1; i <= M; i++) {
    for(int j = 0; j < i; j++) {
      ++G[B[i] - B[j]];
    }
  }
  reverse(begin(G), end(G));
  auto X = FFT::multiply(F, G);
  for(int i = 0; i < Q; i++) {
    if(C[i] == 0) printf("%lld\n", X[200000]);
    else printf("%lld\n", X[200000 - C[i]] + X[200000 + C[i]]);
  }
}