#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const int N = 1e6 + 7;
int main() {
  long long n;
  scanf("%lld", &n);
  long long q[n];
  for (long long i = 1; i < (int)n; i++) scanf("%lld", &q[i]);
  long long a[n + 1];
  a[n] = 1000000;
  for (long long i = n - 1; i >= 1; i--) {
    a[i] = a[i + 1] - q[i];
  }
  long long min = *min_element(a + 1, a + n + 1);
  min--;
  bool flag = true;
  map<long long, long long> m;
  for (long long i = 1; i < (int)n + 1; i++) m[a[i] - min]++;
  for (long long i = 1; i < (int)n + 1; i++) {
    if (m[i] != 1) flag = false;
  }
  if (!flag)
    printf("-1");
  else
    for (long long i = 1; i < (int)n + 1; i++) printf("%lld ", a[i] - min);
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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;
const long long M = 1e9 + 7;
const int N = 1e6 + 7;
int main() {
  long long n;
  scanf("%lld", &n);
  long long q[n];
  for (long long i = 1; i < (int)n; i++) scanf("%lld", &q[i]);
  long long a[n + 1];
  a[n] = 1000000;
  for (long long i = n - 1; i >= 1; i--) {
    a[i] = a[i + 1] - q[i];
  }
  long long min = *min_element(a + 1, a + n + 1);
  min--;
  bool flag = true;
  map<long long, long long> m;
  for (long long i = 1; i < (int)n + 1; i++) m[a[i] - min]++;
  for (long long i = 1; i < (int)n + 1; i++) {
    if (m[i] != 1) flag = false;
  }
  if (!flag)
    printf("-1");
  else
    for (long long i = 1; i < (int)n + 1; i++) printf("%lld ", a[i] - min);
  return 0;
}

```