To improve the boomerang throwing skills of the animals, Zookeeper has set up an n × n grid with some targets, where each row and each column has at most 2 targets each. The rows are numbered from 1 to n from top to bottom, and the columns are numbered from 1 to n from left to right. 

For each column, Zookeeper will throw a boomerang from the bottom of the column (below the grid) upwards. When the boomerang hits any target, it will bounce off, make a 90 degree turn to the right and fly off in a straight line in its new direction. The boomerang can hit multiple targets and does not stop until it leaves the grid.

<image>

In the above example, n=6 and the black crosses are the targets. The boomerang in column 1 (blue arrows) bounces 2 times while the boomerang in column 3 (red arrows) bounces 3 times.

The boomerang in column i hits exactly a_i targets before flying out of the grid. It is known that a_i ≤ 3.

However, Zookeeper has lost the original positions of the targets. Thus, he asks you to construct a valid configuration of targets that matches the number of hits for each column, or tell him that no such configuration exists. If multiple valid configurations exist, you may print any of them.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5).

The next line contains n integers a_1,a_2,…,a_n (0 ≤ a_i ≤ 3).

Output

If no configuration of targets exist, print -1.

Otherwise, on the first line print a single integer t (0 ≤ t ≤ 2n): the number of targets in your configuration. 

Then print t lines with two spaced integers each per line. Each line should contain two integers r and c (1 ≤ r,c ≤ n), where r is the target's row and c is the target's column. All targets should be different. 

Every row and every column in your configuration should have at most two targets each.

Examples

Input


6
2 0 3 0 1 1


Output


5
2 1
2 5
3 3
3 6
5 6


Input


1
0


Output


0


Input


6
3 2 2 2 1 1


Output


-1

Note

For the first test, the answer configuration is the same as in the picture from the statement.

For the second test, the boomerang is not supposed to hit anything, so we can place 0 targets.

For the third test, the following configuration of targets matches the number of hits, but is not allowed as row 3 has 4 targets.

<image>

It can be shown for this test case that no valid configuration of targets will result in the given number of target hits.
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace __gnu_pbds;
using namespace std;

// using long doubles saves you from corner cases but is very time consuming
#define double      	long double
#define int         	long long
#define pb          	push_back
#define pii         	pair<int,int>
#define vi          	vector<int>
#define vii         	vector<pii>
#define mi          	map<int,int>
#define mii         	map<pii,int>
#define all(a)      	(a).begin(),(a).end()
#define sz(x)       	(int)x.size()
// just comment the line below in case of interactive problems
#define endl        	"\n"
#define repp(i,a,b) 	for(int i=a;i<b;i++)
#define rep(i,a,b) 		for(int i=a;i<=b;i++)
#define brep(i,a,b)     for(int i=a;i>=b;i--)
#define deb1(x)      	cout << #x << "=" << x << endl
#define deb2(x, y)  	cout << #x << "=" << x << "," << #y << "=" << y << endl
#define deb3(x, y, z)  	cout << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define trace(v) 		for(auto it=v.begin();it!=v.end();it++)cout<<*it<<" ";cout<<endl;
#define tracearr(a,l,r) for(int iii=l;iii<=r;iii++)cout << a[iii] << " ";cout << endl;
#define PI          	3.1415926535897932384626
#define F 				first
#define S 				second
#define clr(x,y) 		memset(x, y, sizeof(x))
#define IOS         	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);


typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
// *s.find_by_order(1) - gives the 2nd samllest element in set
// s.order_of_key(x)   - gives the number of elements in the set which are strictly smaller than x


const int N=1e5 + 5;
const int MOD=1e9 + 7;

int32_t main()
{
	IOS;
	int T=1;
	//cin >> T;
	// int numOfSetBits = __builtin_popcountll(n);
	std::cout << std::fixed << std::setprecision(12);
	while(T--)
	{
		int n;
		cin >> n;
		int a[n+5];
		rep(i,1,n)
			cin >> a[i];
		int c = 0;
		queue<int> q;
		mi mp;
		rep(i,1,n)
		{
			if(a[i]==1 || a[i]==0)
				continue;
			if(a[i]==3)
			{
				if(!q.empty())
				{
					int f = q.front();
					q.pop();
					mp[f] = i;
				}
				q.push(i);
			}
			else if(a[i]==2)
			{
				if(!q.empty())
				{
					int f = q.front();
					q.pop();
					mp[f] = i;	
				}
			}	
		}
		while(!q.empty())
			q.pop();
		rep(i,1,n)
		{
			if(a[i]==0)
				continue;
			if(a[i]==3)
			{
				if(mp.find(i)==mp.end())
					q.push(i);
			}
			else if(a[i]==2)
			{
				q.push(i);
			}
			else
			{
				if(!q.empty())
				{
					int f = q.front();
					q.pop();
					mp[f] = i;	
				}
			}
		}
		if(!q.empty())
		{
			cout << "-1" << endl;
			return 0;
		}
		int x = 1;
		int f = 0;
		vii ans;
		int done[n+5];
		clr(done,0);
		rep(i,1,n)
		{
			if(a[i]==0)
				continue;
			else if(a[i]==1)
			{
				if(done[i]==0)
				{
					ans.pb({x,i});
					x++;
					done[i] = 1;
				}
			}
			else if(a[i]==2)
			{
				ans.pb({x,i});
				int j = mp[i];
				ans.pb({x,j});
				x++;
				done[j] = 1;
			}
			else
			{
				ans.pb({x,i});
				int j = mp[i];
				ans.pb({x,j});
				if(a[j]==1)
				{
					ans.pb({n-f,j});
					done[j] = 1;
				}
				f++;
				x++;
			}
		}
		cout << ans.size() << endl;
		for(auto it: ans)
			cout << it.F << " " << it.S << endl;
		/*char grid[n+5][n+5];
		clr(grid,' ');
		for(auto it : ans)
			grid[it.F][it.S] = '*';
		rep(i,1,n)
		{
			rep(j,1,n)
				cout << grid[i][j] << " ";
			cout << endl;	
		} */
	}
}

/*
	* while writing bool comp function for sorting put 1 statement in the end which has no if comdition
 	* in case of multiple test cases and N=1e5 don't intialize arrays as global except for graph vector-array
 	* always check whether or not you are doing mod of a negative number
 	* always use 1LL instead of 1
*/