Sunuke received a d-dimensional hypercube with a side length of l1 × ... × ld as a birthday present. Sunuke placed this rectangular parallelepiped so that the range of the i-th coordinate was 0 or more and li or less, and ate the part that satisfied x1 + ... + xd ≤ s. However, xi represents the i-th coordinate. If the volume of the part eaten by Sunuke is V, it can be proved that d! V (V multiplied by the factorial of d) is an integer. Find the remainder of d! V divided by 1,000,000,007.

Constraints

* 2 ≤ d ≤ 300
* 1 ≤ li ≤ 300
* 0 ≤ s ≤ $ \ sum l_ {i} $
* All inputs are integers

Input


d
l1
.. ..
ld
s


Output

Output d! V divided by 1,000,000,007.

Examples

Input

2
6
3
4


Output

15


Input

5
12
34
56
78
90
123


Output

433127538
#include <bits/stdc++.h>
using namespace std;

template <int mod = (int)(1e9 + 7)>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt res(1), mul(x);
    while (n) {
      if (n & 1) res *= mul;
      mul *= mul;
      n >>= 1;
    }
    return res;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};

struct Combination {
  vector<ModInt<>> _fact, _rfact, _inv;
  Combination(long long nsize = 5000000)
      : _fact(nsize + 1), _rfact(nsize + 1), _inv(nsize + 1) {
    _fact[0] = _rfact[nsize] = _inv[0] = 1;
    for (int i = 1; i <= nsize; i++) _fact[i] = _fact[i - 1] * i;
    _rfact[nsize] /= _fact[nsize];
    for (int i = nsize - 1; i >= 0; i--) _rfact[i] = _rfact[i + 1] * (i + 1);
    for (int i = 1; i <= nsize; i++) _inv[i] = _rfact[i] * _fact[i - 1];
  }
  inline ModInt<> fact(int k) const { return _fact[k]; }

  inline ModInt<> rfact(int k) const { return _rfact[k]; }

  inline ModInt<> inv(int k) const { return _inv[k]; }

  ModInt<> P(int n, int r) const {
    if (r < 0 || n < r) return 0;
    return fact(n) * rfact(n - r);
  }

  ModInt<> C(int p, int q) const {
    if (q < 0 || p < q) return 0;
    return fact(p) * rfact(q) * rfact(p - q);
  }
  // n types,choose r
  ModInt<> H(int n, int r) const {
    if (n < 0 || r < 0) return (0);
    return r == 0 ? 1 : C(n + r - 1, r);
  }
};

long long d, s;
vector<long long> l;
vector<vector<vector<ModInt<>>>> dp;

ModInt<> solve();

int main() {
  cin >> d;
  l.resize(d);
  for (int i = 0; i < d; ++i) cin >> l[i];
  cin >> s;
  cout << solve() << endl;
  return 0;
}

ModInt<> solve() {
  ModInt<> res = 0;
  dp.assign(2, vector<vector<ModInt<>>>(s + 1, vector<ModInt<>>(2, 0)));
  dp[0][0][0] = 1;
  for (int i = 0; i < d; ++i)
    for (int j = 0; j <= s; ++j)
      for (int flg = 0; flg < 2; ++flg) {
        dp[(i + 1) & 1][j][flg] = dp[i & 1][j][flg];
        if (j >= l[i]) dp[(i + 1) & 1][j][flg] += dp[i & 1][j - l[i]][1 - flg];
      }
  for (int i = 0; i <= s; ++i)
    res += ModInt<>(s - i).pow(d) * (dp[d & 1][i][0] - dp[d & 1][i][1]);
  return res;
}

