#include <bits/stdc++.h>
using namespace std;
#define N 100010
int i,j,k,m,n,x,t,y,vis[N],a[N];
char s[N];
int main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (i = 1; i <= n; ++i)vis[i] = (s[i] == '1');
    if (!vis[1] || !vis[n - 1] || vis[n])return puts("-1"), 0;
    for (i = 2; i < n - 1; ++i)if (vis[i] != vis[n - i])return puts("-1"), 0;
    for (x = 0, i = 1; i <= n; ++i) {
        if (vis[i])while (x > 0)printf("%d %d\n", a[x--], i);
        if (i != n)a[++x] = i;
    }
    while (x > 0)printf("%d %d\n", a[x--], n);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given a string s of length n. 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.
* If the i-th character in s is `1`, we can have a connected component of size i by removing one edge from the tree.
* If the i-th character in s is `0`, we cannot have a connected component of size i by removing any one edge from the tree.



If such a tree exists, construct one such tree.

Constraints

* 2 \leq n \leq 10^5
* s is a string of length n consisting of `0` and `1`.

Input

Input is given from Standard Input in the following format:


s


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

1111


Output

-1


Input

1110


Output

1 2
2 3
3 4


Input

1010


Output

1 2
1 3
1 4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define N 100010
int i,j,k,m,n,x,t,y,vis[N],a[N];
char s[N];
int main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (i = 1; i <= n; ++i)vis[i] = (s[i] == '1');
    if (!vis[1] || !vis[n - 1] || vis[n])return puts("-1"), 0;
    for (i = 2; i < n - 1; ++i)if (vis[i] != vis[n - i])return puts("-1"), 0;
    for (x = 0, i = 1; i <= n; ++i) {
        if (vis[i])while (x > 0)printf("%d %d\n", a[x--], i);
        if (i != n)a[++x] = i;
    }
    while (x > 0)printf("%d %d\n", a[x--], n);
}

```