You are given a sequence D_1, D_2, ..., D_N of length N. The values of D_i are all distinct. Does a tree with N vertices that satisfies the following conditions exist?

* The vertices are numbered 1,2,..., N.
* The edges are numbered 1,2,..., N-1, and Edge i connects Vertex u_i and v_i.
* For each vertex i, the sum of the distances from i to the other vertices is D_i, assuming that the length of each edge is 1.



If such a tree exists, construct one such tree.

Constraints

* 2 \leq N \leq 100000
* 1 \leq D_i \leq 10^{12}
* D_i are all distinct.

Input

Input is given from Standard Input in the following format:


N
D_1
D_2
:
D_N


Output

If a tree with n vertices that satisfies the conditions does not exist, print `-1`.

If a tree with n vertices that satisfies the conditions exist, print n-1 lines. The i-th line should contain u_i and v_i with a space in between. If there are multiple trees that satisfy the conditions, any such tree will be accepted.

Examples

Input

7
10
15
13
18
11
14
19


Output

1 2
1 3
1 5
3 4
5 6
6 7


Input

2
1
2


Output

-1


Input

15
57
62
47
45
42
74
90
75
54
50
66
63
77
87
51


Output

1 10
1 11
2 8
2 15
3 5
3 9
4 5
4 10
5 15
6 12
6 14
7 13
9 12
11 13
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <queue>
#define fi first
#define se second
using namespace std;
typedef long long i64;
typedef pair<i64, int> pii;
inline i64 read(int f = 1, i64 x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5;
int n, s[N], u[N], v[N]; pii p[N]; i64 sum;
int main()
{
    n = read(); for(int i = 1; i <= n; ++i) p[i].fi = read(), p[i].se = i, s[i] = 1;
    sort(p+1, p+n+1); 
    for(int i = n; i > 1; --i)
    {
        i64 w = p[i].fi-n+s[i]*2; int j = lower_bound(p+1, p+1+n, pii(w, 0))-p;
        if(p[j].fi != w) return puts("-1"), 0; 
        u[i] = p[i].se, v[i] = p[j].se, s[j] += s[i], sum += s[i];
    }
    if(sum != p[1].fi) return puts("-1"), 0;
    for(int i = 2; i <= n; ++i) printf("%d %d\n", u[i], v[i]);
    return 0;
}