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>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef string str;

#define pb push_back
#define ppp pop_back
#define pii pair<int,int>
#define fi first
#define se second
#define stie std::tie
#define vec vector
#define	forn(i, l, r) for (int i=l; i<=r; i++)
#define forb(i, r, l) for (int i=r; i>=l; i--)
#define emp empty
#define beg begin
#define ins insert
#define cle clear
#define era erase
#define que queue
#define pque priority_queue
#define mset multiset
#define deq deque
#define sta stack
#define con const
#define rsz resize
#define ass assign
#define lowb lower_bound
#define uppb upper_bound

template<class T> void mini(T& a, T b) { a = min(a, b); }
template<class T> void maxi(T& a, T b) { a = max(a, b); }
template<class T1, class T2> ostream& operator<<(ostream &out, pair<T1, T2> a) { return out << a.first << " " << a.second; }
template<class T> ostream& operator<<(ostream& out, vector<T> a) { for (auto& i : a) out << i << " "; return out; }
template<class T> ostream& operator<<(ostream& out, set<T> a) { for (auto& i : a) out << i << " "; return out; }
template<class T> ostream& operator<<(ostream& out, multiset<T> a) { for (auto& i : a) out << i << " "; return out; }
template<class T1, class T2> ostream& operator<<(ostream& out, map<T1, T2> a) { for (auto& i : a) out << i << "\n"; return out; }
template<class T1, class T2> ostream& operator<<(ostream& out, unordered_map<T1, T2> a) { for (auto& i : a) out << i << "\n"; return out; }
template<class T> ostream& operator<<(ostream& out, queue<T> a) { while (!a.empty()) { out << a.front() << " "; a.pop(); } return out; }
template<class T> ostream& operator<<(ostream& out, deque<T> a) { while (!a.empty()) { out << a.front() << " "; a.pop_front(); } return out; }
template<class T> ostream& operator<<(ostream& out, priority_queue<T> a) { while (!a.empty()) { out << a.top() << "\n"; a.pop(); } return out; }
template<class T> void out(T a) { cout << a << endl; }
template<class T1, class T2> void out(T1 a, T2 b) { cout << a << " " << b << endl; }
template<class T1, class T2, class T3> void out(T1 a, T2 b, T3 c) { cout << a << " " << b << " " << c << endl; }
template<class T1, class T2, class T3, class T4> void out(T1 a, T2 b, T3 c, T4 d) { cout << a << " " << b << " " << c << " " << d << endl; }
template<class T> void out(vector<vector<T>> a) { for (vector<T> i : a) out(i); }
template<class T> void out_(T a[], int l, int r) { for (int i = l; i <= r; i++) cout << a[i] << " "; cout << endl; }
void out() { cout << "OK" << endl; }
template<class T> void sort(vector<T>& a) { sort(a.begin(), a.end()); }
template<class T> void uniq(vector<T>& a) { sort(a); a.erase(unique(a.begin(), a.end()), a.end()); }
template<class T> vector<T> set_int(vector<T> a, vector<T> b) { sort(a); sort(b); vector<T> res; set_intersection(a.begin(), a.end(), b.begin(), b.end(), back_inserter(res)); return res; }
clock_t start_time; void start_timer() { start_time = clock(); }
double get_time() { return (double)(clock() - start_time) / CLOCKS_PER_SEC; }
#if __SIZEOF_INT128__ >= 16
typedef __int128 LL;
istream& operator>>(istream& in, __int128& a) { int64_t b; cin >> b; a = b; return in; }
ostream& operator<<(ostream& out, const __int128 a) {
	unsigned __int128 b = a < 0 ? -a : a; char buf[128]; char* c = end(buf); do { --c; *c = "0123456789"[b % 10]; b /= 10; } while (b);
	if (a < 0) { --c; *c = '-'; } int len = end(buf) - c; out.rdbuf()->sputn(c, len); return out;
}
#endif

con int N = 1e6 + 5;
con int K = 505;

int c[N];
vec<pii> es;
int sz;
int up[K];
bool used[N];

struct E {
	int u, v, w;
};

bool operator<(con E& a, con E& b) {
	return a.w < b.w;
};

vec<E> all;

int dep[K];
int par[K];
int val[K];

void init(int n) {
	forn(i, 1, n) {
		dep[i] = 0;
		par[i] = i;
		val[i] = 0;
	}
}

int find(int v) {
	if (par[v] == v) return v;
	return par[v] = find(par[v]);
}

void unite(int u, int v, int w) {
	u = find(u);
	v = find(v);
	if (u == v) return;
	if (dep[u] < dep[v]) swap(u, v);
	par[v] = u;
	val[u] = w;
	if (dep[u] == dep[v]) dep[u]++;
}

map<int, int> Map;

void Solve() {
	int n;
	cin >> n;
	forn(i, 1, n) {
		forn(j, 1, n) {
			int x;
			cin >> x;
			if (i == j) c[i] = x;
			else if (i < j) {
				all.pb({i, j, x});
			}
		}
	}
	sz = n;
	sort(all);
	forn(i, 1, n) up[i] = i;
	int l = 0;
	while (l < all.size()) {
		int r = l;
		while (r + 1 < all.size() && all[r + 1].w == all[l].w) r++;
		init(n);
		forn(i, l, r) {
			unite(all[i].u, all[i].v, all[i].w);
		}
		Map.cle();
		forn(i, 1, n) {
			int v = find(i);
			if (dep[v] == 0) continue;
			if (!Map.count(v)) {
				Map[v] = ++sz;
				c[sz] = val[v];
			}
			int kek = Map[v];
			if (!used[up[i]]) es.pb({up[i], kek});
			used[up[i]] = 1;
			up[i] = kek;
		}
		l = r + 1;
	}
	cout << sz << "\n";
	forn(i, 1, sz) {
		cout << c[i] << " ";
	}
	cout << "\n";
	cout << sz << "\n";
	for (auto [u, v] : es) {
		cout << u << " " << v << "\n";
	}
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
#endif
	start_timer();
	Solve();
	return 0;
}