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;
int const MAX = 100 * 100 + 41;
int const MOD = 1000 * 1000 * 1000 + 7;
int n;
int a[MAX];
int inva[MAX];
int d[2][MAX];
int pref[MAX];
int ans[MAX];
int add(int a, int b) { return (a + b) % MOD; }
int mult(int a, int b) { return (int)((long long)a * b % MOD); }
int sub(int a, int b) {
  a -= b;
  while (a < 0) a += MOD;
  return a;
}
int bp(int x, int d) {
  int res = 1;
  while (d) {
    if (d & 1) res = mult(res, x);
    x = mult(x, x);
    d >>= 1;
  }
  return res;
}
int inv(int x) { return bp(x, MOD - 2); }
void init() {
  pref[0] = 1;
  for (int i = 1; i <= n - 1; i++) {
    pref[i] = mult(pref[i - 1], a[i]);
  }
  for (int i = 0; i <= n; i++) {
    inva[i] = inv(sub(a[i], 1));
  }
}
void solvediff() {
  memset(d, 0, sizeof(d));
  for (int i = 1; i <= n - 1; i++) {
    for (int j = i + 1; j <= n; j++) {
      ans[j - i] = add(ans[j - i], pref[j - 1]);
    }
  }
  for (int i = n; i >= 1; i--) {
    memset(d[1], 0, sizeof(d[1]));
    int cur = mult(a[i], a[i] - 1);
    for (int j = i + 2; j <= n; j++) {
      cur = mult(cur, a[j - 1]);
      d[1][j - i + 1] = add(d[1][j - i + 1], cur);
    }
    for (int j = 0; j <= n * 2 - 2; j++) {
      if (d[0][j]) {
        int v1 = d[0][j];
        int v2 = mult(a[i], a[i] - 1);
        int v3 = mult(a[i + 1], inva[i + 1]);
        int v = mult(v1, mult(v2, v3));
        d[1][j + 2] = add(d[1][j + 2], v);
      }
    }
    for (int j = 0; j <= n * 2 - 2; j++) {
      ans[j] = add(ans[j], mult(pref[i - 1], d[1][j]));
    }
    memcpy(d[0], d[1], sizeof(d[0]));
  }
}
void solvesame() {
  int ip2 = inv(2);
  for (int i = 1; i <= n - 1; i++) {
    int cur = mult(a[i], a[i] - 1);
    cur = mult(cur, pref[i - 1]);
    for (int j = i + 1; j <= n; j++) {
      int h = j - i + j - i;
      ans[h] = add(ans[h], mult(cur, ip2));
      cur = mult(cur, a[j]);
      cur = mult(cur, a[j]);
    }
  }
}
void solve() {
  init();
  solvediff();
  solvesame();
  for (int i = 1; i <= n * 2 - 2; i++) {
    printf("%d ", ans[i]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d", &a[i]);
  }
  solve();
  return 0;
}