#include <bits/stdc++.h>
using namespace std;
const int maxn = 4e5 + 5;
const int MOD = 1e9 + 7;
int d[maxn];
int a[maxn];
map<int, int> fl;
int main() {
  int n = 0;
  while (~scanf("%d", &n)) {
    for (int i = 1; i < n; i++) {
      scanf("%d", d + i);
    }
    int left = 1;
    int right = n;
    int mid;
    bool f = 0;
    while (right >= left) {
      mid = (left + right) / 2;
      a[1] = mid;
      fl.clear();
      fl[a[1]] = 1;
      int tmp = 0;
      int flag = 0;
      for (int i = 1; i < n; i++) {
        a[i + 1] = a[i] + d[i];
        if (a[i + 1] > n || a[i + 1] < 1) {
          tmp = i + 1;
          flag = 1;
          break;
        }
        if (fl[a[i + 1]] != 0) {
          flag = 1;
          f = 1;
          break;
        }
        fl[a[i + 1]] = 1;
      }
      if (flag == 0) {
        break;
      }
      if (a[tmp] < 1) {
        left = mid + 1;
      } else if (a[tmp] > n) {
        right = mid - 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] > n || a[i] < 1) {
        f = 1;
        break;
      }
    }
    if (f == 1) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      printf("%d%c", a[i], i == n ? '\n' : ' ');
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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 int maxn = 4e5 + 5;
const int MOD = 1e9 + 7;
int d[maxn];
int a[maxn];
map<int, int> fl;
int main() {
  int n = 0;
  while (~scanf("%d", &n)) {
    for (int i = 1; i < n; i++) {
      scanf("%d", d + i);
    }
    int left = 1;
    int right = n;
    int mid;
    bool f = 0;
    while (right >= left) {
      mid = (left + right) / 2;
      a[1] = mid;
      fl.clear();
      fl[a[1]] = 1;
      int tmp = 0;
      int flag = 0;
      for (int i = 1; i < n; i++) {
        a[i + 1] = a[i] + d[i];
        if (a[i + 1] > n || a[i + 1] < 1) {
          tmp = i + 1;
          flag = 1;
          break;
        }
        if (fl[a[i + 1]] != 0) {
          flag = 1;
          f = 1;
          break;
        }
        fl[a[i + 1]] = 1;
      }
      if (flag == 0) {
        break;
      }
      if (a[tmp] < 1) {
        left = mid + 1;
      } else if (a[tmp] > n) {
        right = mid - 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] > n || a[i] < 1) {
        f = 1;
        break;
      }
    }
    if (f == 1) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      printf("%d%c", a[i], i == n ? '\n' : ' ');
    }
  }
  return 0;
}

```