Problem: The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.

The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.

Input

The first line contains a single integer n (2 ≤ n ≤ 500) — the number of lower-level employees.

This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 ≤ a_{i,j} ≤ 5000) — salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.

Output

In the first line, print a single integer k — the number of employees in the company.

In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.

In the third line, print a single integer r — the number of the employee who is the head of the company.

In the following k-1 lines, print two integers v and u (1 ≤ v, u ≤ k) — the number of the employee and his direct supervisor.

Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.

Example

Input


3
2 5 7
5 1 7
7 7 4


Output


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

Note

One of the possible structures in the first example: <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define forr(i,a,b) for(int i=(a);i<(b);i++)
#define forn(i,n) forr(i,0,n)
#define dforn(i,n) for(int i=n-1;i>=0;i--)
#define forall(it,v) for(auto it=v.begin();it!=v.end();it++)
#define sz(c) ((int)c.size())
#define rsz resize
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define fst first
#define snd second

#ifdef ANARAP
#define MAXN 13
#else
#define MAXN 502
#endif

using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

int mx[MAXN][MAXN];
bool used[MAXN][MAXN];
struct UnionFind{
	vector<int> f, setSize; //the array f contains the parent of each node
	int cantSets;
	void init(int n)
	{
		f.clear(); setSize.clear();
		cantSets = n;
		f.rsz(n, -1);
		setSize.rsz(n, 1);
	}
	int comp(int x) {return (f[x]==-1? x : f[x]=comp(f[x]));}//O(1)
	bool join(int a, int b) //devuelve true si ya estaban juntos
	{
		a=comp(a), b= comp(b);
		if(a!=b){
			setSize[b] += setSize[a];
			f[a] = b; //el grupo mas grande (b) ahora representa al mas chico (a)
		}
		return a == b;
	}
}uf;

int main()
{
	#ifdef ANARAP
		freopen("input.in", "r", stdin);
		//freopen("output.out", "w", stdout);
		unsigned t0, t1; t0=clock();
	#endif
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	
	int n; cin >> n;
	vector<int> ans(n), apunta(n, -1);
	forn(i,n) forn(j,n) cin >> mx[i][j];
	forn(i,n) ans[i] = mx[i][i];
	int row = -1;
	int mini = 5001;
	forn(i,n)forr(j,i+1,n)if(mx[i][j]<mini) {row = i; mini = mx[i][j];}
	int node=n;
	uf.init(n*2+10);
	vector<int> rows;
	forr(j, row+1, n){
		if(mx[row][j] == mini){
			used[row][j] = 1;
			rows.pb(j);
			uf.join(j, node);
			apunta[j] = node;
		} 
	}
	forn(i,sz(rows)){
		forr(j, rows[i]+1, n){
			if(mx[rows[i]][j] == mini) used[rows[i]][j]=1;
		}
	}
	apunta[row] = node;
	uf.join(row, node);
	ans.pb(mini);
	apunta.pb(-1);
	while(1){
		node++;
		row = -1;
		mini = 5001;
		rows.clear();
		forn(i,n)forr(j,i+1,n)if(!used[i][j] && mx[i][j]<mini) {row = i; mini = mx[i][j];}
		//~ forn(i,n){
			//~ forn(j,n) cout << used[i][j] << " ";
			//~ cout << '\n';
		//~ }
		//~ cout << row << endl;
		if(row==-1) break;
		ans.pb(mini);
		apunta.pb(-1);
		int auxrow = uf.comp(row);
		forn(i,n)if(uf.comp(i) == auxrow) {rows.pb(i);}
		apunta[auxrow] = node;
		uf.join(auxrow, node);
		vector<int> rows2;
		forn(i, sz(rows)){
			forr(j, rows[i], n){
				if(!used[rows[i]][j] && mx[rows[i]][j] == mini){
					used[rows[i]][j] = 1;
					rows2.pb(j);
					int aux = uf.comp(j);
					if(aux!=node){
						apunta[aux] = node;
						uf.join(aux, node);
					}
				} 
			}
		}
		forn(i, sz(rows2)){
			forr(j, rows2[i], n){
				if(!used[rows2[i]][j] && mx[rows2[i]][j] == mini){
					used[rows2[i]][j] = 1;
					int aux = uf.comp(j);
					if(aux!=node){
						apunta[aux] = node;
						uf.join(aux, node);
					}
				} 
			}
		}
	}
	cout << node << '\n';
	forn(i,node) cout << ans[i] << " ";
	cout << '\n';
	int root = -1;
	forn(i,node) if(apunta[i] == -1) root = i;
	cout << root+1 << '\n';
	forn(i,node){
		if(apunta[i]!=-1) cout << i+1 << " " << apunta[i]+1 << '\n';
		
	}
	
	
	
	#ifdef ANARAP
	t1 = clock();double time = (double(t1-t0)/CLOCKS_PER_SEC); cout << "Execution Time: " << time << endl;
	#endif
	return 0;
}