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:
/**
 * Author: Daniel
 * Created Time: 2021-03-13 13:41:37
**/

// time-limit: 2000
#include <bits/stdc++.h>

using namespace std;

#define F first 
#define S second
#define ER erase
#define IS insert
#define PI acos(-1)
#define PB pop_back
#define EB emplace_back
#define lowbit(x) (x & -x)
#define SZ(x) ((int)x.size())
#define MP(x, y) make_pair(x, y)
#define ALL(x) x.begin(), x.end()
#define RALL(x) x.rbegin(), x.rend()
#define SOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout<<fixed<<setprecision(10)

typedef long long LL;
typedef vector<LL> VL;
typedef vector<int> VI;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef unsigned long long ULL;
typedef pair<double, double> PDD;
typedef vector<pair<int, int> > VPII;

template <typename A> using VE = vector<A>;
template <typename A> using USET = unordered_set<A>;
template <typename A> using HEAP = priority_queue<A>;
template <typename A, typename B> using PA = pair<A, B>;
template <typename A, typename B> using UMAP = unordered_map<A, B>;
template <typename A> using RHEAP = priority_queue<A, vector<A>, greater<A> >;

template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) { return '"' + s + '"'; }

string to_string(const char* s) { return to_string((string)s); } 

string to_string(const char c) { return to_string((string)"" + c); } 

string to_string(bool b) { return (b ? "true" : "false"); }

string to_string(vector<bool> v) 
{
    bool first = true;
    string res = "{";
    for (int i = 0; i < static_cast<int>(v.size()); i++) 
    {
        if (!first) res += ", ";
        first = false;
        res += to_string(v[i]);
    }
    res += "}";
    return res;
}

template <size_t N>
string to_string(bitset<N> v)
{
    string res = "";
    for (size_t i = 0; i < N; i++) res += static_cast<char>('0' + v[i]);
    return res;
}

template <typename A>
string to_string(A v)
{
    bool first = true;
    string res = "{";
    for (const auto &x : v)
    {
        if (!first) res += ", ";
        first = false;
        res += to_string(x);
    }
    res += "}";
    return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) { return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")"; }

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) { return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")"; }

void debug_out() { cout << '\n'; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cout << " " << to_string(H); debug_out(T...); }

#ifdef LOCAL
#define debug(...) cout << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////

// check the limitation!!!
const int N = 5010, M = 1010;

int n;
int a[N][N];
int res[N];
int p[N];
struct Node
{
    int u, v, val;
    bool operator<(const Node &t) const
    {
        if (val != t.val) return val < t.val;
        else return u < t.u;
    }
};

int find(int x)
{
    if (x != p[x]) p[x] = find(p[x]);
    return p[x];
}

// read the question carefully!!!
int main()
{
    SOS;

    cin >> n;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            cin >> a[i][j];
    VE<Node> b;
    for (int i = 1; i <= n; i ++ )
        for (int j = i; j <= n; j ++ )
            b.EB((Node){i, j, a[i][j]});
    sort(ALL(b));
    VPII edge;
    int m = n;
    for (int i = 1; i <= n; i ++ ) 
    {
        p[i] = i;
        res[i] = a[i][i];
    }
    for (auto [u, v, val] : b)
    {
        int pa = find(u), pb = find(v);
        if (pa == pb) continue;
        else if (val == res[pb])
        {
            p[pa] = pb;
            edge.EB(pa, pb);
        }
        else if (val == res[pa])
        {
            p[pb] = pa;
            edge.EB(pb, pa);
        }
        else
        {
            p[pa] = p[pb] = ++ m;
            p[m] = m;
            res[m] = val;
            edge.EB(pa, m);
            edge.EB(pb, m);
        }
    }
    cout << m << '\n';
    for (int i = 1; i <= m; i ++ )
    {
        cout << res[i];
        cout << " \n"[i == m];
    }
    cout << find(1) << '\n';
    for (auto [x, y] : edge) cout << x << ' ' << y << '\n';
    return 0;
}

// GOOD LUCK!!!