Problem: You are given a rooted tree. Let's denote d(x) as depth of node x: depth of the root is 1, depth of any other node x is d(y) + 1, where y is a parent of x.

The tree has the following property: every node x with d(x) = i has exactly ai children. Maximum possible depth of a node is n, and an = 0.

We define fk as the number of unordered pairs of vertices in the tree such that the number of edges on the simple path between them is equal to k.

Calculate fk modulo 109 + 7 for every 1 ≤ k ≤ 2n - 2.

Input

The first line of input contains an integer n (2 ≤ n ≤ 5 000) — the maximum depth of a node.

The second line of input contains n - 1 integers a1, a2, ..., an - 1 (2 ≤ ai ≤ 109), where ai is the number of children of every node x such that d(x) = i. Since an = 0, it is not given in the input.

Output

Print 2n - 2 numbers. The k-th of these numbers must be equal to fk modulo 109 + 7.

Examples

Input

4
2 2 2


Output

14 19 20 20 16 16 

Input

3
2 3


Output

8 13 6 9 

Note

This the tree from the first sample: 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
const int MOD = 1000000007;
struct Mint {
  int val;
  Mint(long long v = 0) {
    if (v < 0) v = v % MOD + MOD;
    if (v >= MOD) v %= MOD;
    val = v;
  }
  static int mod_inv(int a, int m = MOD) {
    int g = m, r = a, x = 0, y = 1;
    while (r != 0) {
      int q = g / r;
      g %= r;
      swap(g, r);
      x -= q * y;
      swap(x, y);
    }
    return x < 0 ? x + m : x;
  }
  explicit operator int() const { return val; }
  Mint &operator+=(const Mint &other) {
    val += other.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  Mint &operator-=(const Mint &other) {
    val -= other.val;
    if (val < 0) val += MOD;
    return *this;
  }
  static unsigned fast_mod(uint64_t x, unsigned m = MOD) {
    return x % m;
    unsigned x_high = x >> 32, x_low = (unsigned)x;
    unsigned quot, rem;
    asm("divl %4\n" : "=a"(quot), "=d"(rem) : "d"(x_high), "a"(x_low), "r"(m));
    return rem;
  }
  Mint &operator*=(const Mint &other) {
    val = fast_mod((uint64_t)val * other.val);
    return *this;
  }
  Mint &operator/=(const Mint &other) { return *this *= other.inv(); }
  friend Mint operator+(const Mint &a, const Mint &b) { return Mint(a) += b; }
  friend Mint operator-(const Mint &a, const Mint &b) { return Mint(a) -= b; }
  friend Mint operator*(const Mint &a, const Mint &b) { return Mint(a) *= b; }
  friend Mint operator/(const Mint &a, const Mint &b) { return Mint(a) /= b; }
  Mint &operator++() {
    val = val == MOD - 1 ? 0 : val + 1;
    return *this;
  }
  Mint &operator--() {
    val = val == 0 ? MOD - 1 : val - 1;
    return *this;
  }
  Mint operator++(int32_t) {
    Mint before = *this;
    ++*this;
    return before;
  }
  Mint operator--(int32_t) {
    Mint before = *this;
    --*this;
    return before;
  }
  Mint operator-() const { return val == 0 ? 0 : MOD - val; }
  bool operator==(const Mint &other) const { return val == other.val; }
  bool operator!=(const Mint &other) const { return val != other.val; }
  Mint inv() const { return mod_inv(val); }
  Mint power(long long p) const {
    assert(p >= 0);
    Mint a = *this, result = 1;
    while (p > 0) {
      if (p & 1) result *= a;
      a *= a;
      p >>= 1;
    }
    return result;
  }
  friend ostream &operator<<(ostream &stream, const Mint &m) {
    return stream << m.val;
  }
  friend istream &operator>>(istream &stream, Mint &m) {
    return stream >> m.val;
  }
};
const int N = 5005;
Mint dp[N][2 * N];
Mint a[N], b[N];
Mint ans[2 * N];
int _runtimeTerror_() {
  int n;
  cin >> n;
  a[0] = 1;
  for (int i = 1; i < n; ++i) cin >> a[i];
  a[n] = 0;
  b[0] = a[0];
  for (int i = 1; i < n; ++i) b[i] = b[i - 1] * a[i];
  for (int i = 1; i <= n; ++i) {
    Mint val = a[i] - 1, x = 1, y = 1;
    if (i != n) {
      dp[i][1] += a[i] - 1;
      ans[1] += x * b[i - 1];
    }
    for (int j = i + 1; j <= n - 1; ++j) {
      val *= a[j];
      x *= a[j];
      y *= a[j];
      ans[j - i + 1] += x * b[i - 1];
      dp[i][j - i + 1] += val;
    }
    if (i != 1) ++dp[i][1];
    for (int j = 1; j <= 2 * n; ++j) dp[i][j + 1] += dp[i - 1][j];
  }
  for (int i = 1; i <= 2 * n - 2; ++i) {
    Mint p = 0;
    for (int j = 1; j <= n; ++j) p += b[j - 1] * dp[j][i];
    cout << (p + ans[i]) / 2 << " ";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}