#include <bits/stdc++.h>
using namespace std;
template <class T>
bool ckmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
void read(vector<long long int> &a, int n) {
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
}
bool fine = 0;
vector<long long int> a;
long long int n, m, k, x;
void solve() {
  cin >> n;
  read(a, n - 1);
  vector<long long int> pref(n - 1);
  pref[0] = a[0];
  long long int mx = pref[0];
  for (int i = 1; i <= n - 2; i++) {
    pref[i] = pref[i - 1] + a[i];
    ckmax(mx, pref[i]);
  }
  vector<long long int> ans(n);
  ans[0] = (mx <= 0 ? n : n - mx);
  if (ans[0] <= 0 || ans[0] > n) {
    cout << -1 << "\n";
    return;
  }
  for (int i = 1; i <= n - 1; i++) {
    ans[i] = ans[0] + pref[i - 1];
    if (ans[i] <= 0 || ans[i] > n) {
      cout << -1 << "\n";
      return;
    }
  }
  set<long long int> st(ans.begin(), ans.end());
  if (st.size() < n) {
    cout << -1 << "\n";
    return;
  }
  for (auto t : ans) cout << t << " ";
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i = 1;
  if (fine) {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  } else
    solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

An array of integers p_1, p_2, ..., p_n is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3, 1, 2], [1], [1, 2, 3, 4, 5] and [4, 3, 1, 2]. The following arrays are not permutations: [2], [1, 1], [2, 3, 4].

Polycarp invented a really cool permutation p_1, p_2, ..., p_n of length n. It is very disappointing, but he forgot this permutation. He only remembers the array q_1, q_2, ..., q_{n-1} of length n-1, where q_i=p_{i+1}-p_i.

Given n and q=q_1, q_2, ..., q_{n-1}, help Polycarp restore the invented permutation.

Input

The first line contains the integer n (2 ≤ n ≤ 2⋅10^5) — the length of the permutation to restore. The second line contains n-1 integers q_1, q_2, ..., q_{n-1} (-n < q_i < n).

Output

Print the integer -1 if there is no such permutation of length n which corresponds to the given array q. Otherwise, if it exists, print p_1, p_2, ..., p_n. Print any such permutation if there are many of them.

Examples

Input


3
-2 1


Output


3 1 2 

Input


5
1 1 1 1


Output


1 2 3 4 5 

Input


4
-1 2 2


Output


-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool ckmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
void read(vector<long long int> &a, int n) {
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
}
bool fine = 0;
vector<long long int> a;
long long int n, m, k, x;
void solve() {
  cin >> n;
  read(a, n - 1);
  vector<long long int> pref(n - 1);
  pref[0] = a[0];
  long long int mx = pref[0];
  for (int i = 1; i <= n - 2; i++) {
    pref[i] = pref[i - 1] + a[i];
    ckmax(mx, pref[i]);
  }
  vector<long long int> ans(n);
  ans[0] = (mx <= 0 ? n : n - mx);
  if (ans[0] <= 0 || ans[0] > n) {
    cout << -1 << "\n";
    return;
  }
  for (int i = 1; i <= n - 1; i++) {
    ans[i] = ans[0] + pref[i - 1];
    if (ans[i] <= 0 || ans[i] > n) {
      cout << -1 << "\n";
      return;
    }
  }
  set<long long int> st(ans.begin(), ans.end());
  if (st.size() < n) {
    cout << -1 << "\n";
    return;
  }
  for (auto t : ans) cout << t << " ";
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i = 1;
  if (fine) {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  } else
    solve();
  return 0;
}

```