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 <stdio.h>
#include <iostream>
#include <vector>
#include <unordered_set>
#include <math.h>
#include <time.h>
#include <algorithm>
#include <string.h>
#include <string>
#include <deque>
#include <ctype.h>
#include <limits.h>
#include <map>
#include <set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <functional>
#include <limits.h>
#include <random>

using namespace std;
typedef long long ll;

void solve();

mt19937 rng(time(0));

ll qpow(ll x, ll p, ll MOD) {
    ll res = 1, v = x;
    while (p) {
        if(p & 1) {
            res *= v;
            res %= MOD;
        }
        v *= v;
        v %= MOD;
        p >>= 1;
    }
    return res;
}

ll inv(ll x, ll MOD) {
    return qpow(x, MOD - 2, MOD);
}

void init() {
#ifdef _LOCAL_ 
    if(!freopen("case.txt", "r", stdin)) {
        perror("open file failed!:");
        exit(1);
    }
    if(!freopen("main.ans", "w", stdout)) {
        perror("open file failed!:");
        exit(1);
    }
#endif
}
 
#ifdef _LOCAL_
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#   define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#   define dbg(...)
#endif

ll read() {
    int ch = getchar();
    if(ch == EOF) {
        return INT_MIN;
    }
    while(!isdigit(ch) && ch != '-') {
        ch = getchar();
        if(ch == EOF) {
            return INT_MIN;
        }
    }
    ll res = 0, sign = 1;
    if(ch == '-') {
        ch = getchar();
        sign = -1;
    }
    while(isdigit(ch)) {
        res = res * 10 + ch - '0';
        ch = getchar();
    }
    return res * sign;
}

int main() {
    init();
    solve();
    return 0;
}
/** 写之前 : 
 *  1. MAXN 是否定义正确？
 */
/*********************code start here*********************/
int a[505][505];
int sa[1100] = {0}, fa[1100];
void solve() {
    int n = read();
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            a[i][j] = read();
        }
    }
    memset(fa, -1, sizeof(fa));
    int sz = n;
    for(int i = 0; i < n; ++i) {
        sa[i] = a[i][i];
        for(int j = i+1; j < n; ++j) {
            if(fa[j] == -1) {
                int p = i;
                while(fa[p] != -1 && sa[fa[p]] < a[i][j]) {
                    p = fa[p];
                }
                if(fa[p] == -1) {
                    fa[p] = sz++;
                    sa[fa[p]] = a[i][j];
                }
                if(sa[fa[p]] == a[i][j]) {
                    fa[j] = fa[p];
                } else {
                    int up = fa[p];
                    fa[j] = fa[p] = sz++;
                    fa[fa[p]] = up;
                    sa[fa[p]] = a[i][j];
                }
            } else {
                int ti = i, tj = j;
                while(sa[fa[ti]] < a[i][j]) {
                    ti = fa[ti];
                }
                while(sa[fa[tj]] < a[i][j]) {
                    tj = fa[tj];
                }
                if(sa[fa[ti]] == a[i][j]) {
                    fa[tj] = fa[ti];
                } else if(sa[fa[tj]] == a[i][j]) {
                    fa[ti] = fa[tj];
                } else {
                    int up = fa[ti];
                    fa[ti] = fa[tj] = sz++;
                    fa[fa[ti]] = up;
                    sa[fa[ti]] = a[i][j];
                }
            }
        }
    }
    printf("%d\n", sz);
    for(int i = 0; i < sz; ++i) {
        printf("%d ", sa[i]);
    }
    printf("\n");
    for(int i = 0; i < sz; ++i) {
        if(fa[i] == -1) {
            printf("%d\n", i + 1);
            break;
        }
    }
    for(int i = 0; i < sz; ++i) {
        if(fa[i] != -1) {
            printf("%d %d\n", i + 1, fa[i] + 1);
        }
    }
}

/** 提交前检查 : 
 *  1. 答案和中间结果是否应该是 ll?
 *  2. 是否还有没有读入的数据？
 *  3. 是否有 n = 1 的情况？
 */