Young Table(暴力，交换位置)

 Young Table

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 237B

Appoint description: 
System Crawler  (2016-04-26)

Description

You've got table a, consisting of n rows, numbered from 1 to n. The i-th line of table a contains c_i cells, at that for all i(1 < i ≤ n)holds c_i ≤ c_i - 1.

Let's denote s as the total number of cells of table a, that is, . We know that each cell of the table contains a single integer from 1 to s, at that all written integers are distinct.

Let's assume that the cells of the i-th row of table a are numbered from 1 to c_i, then let's denote the number written in the j-th cell of thei-th row as a_i, j. Your task is to perform several swap operations to rearrange the numbers in the table so as to fulfill the following conditions:

1. for all i, j(1 < i ≤ n; 1 ≤ j ≤ c_i) holds a_i, j > a_{i - 1, j};
2. for all i, j(1 ≤ i ≤ n; 1 < j ≤ c_i) holds a_i, j > a_{i, j - 1}.

In one swap operation you are allowed to choose two different cells of the table and swap the recorded there numbers, that is the number that was recorded in the first of the selected cells before the swap, is written in the second cell after it. Similarly, the number that was recorded in the second of the selected cells, is written in the first cell after the swap.

Rearrange the numbers in the required manner. Note that you are allowed to perform any number of operations, but not more than s. You do not have to minimize the number of operations.

Input

The first line contains a single integer n(1 ≤ n ≤ 50) that shows the number of rows in the table. The second line contains n space-separated integers c_i(1 ≤ c_i ≤ 50; c_i ≤ c_i - 1) — the numbers of cells on the corresponding rows.

Next n lines contain table а. The i-th of them contains c_i space-separated integers: the j-th integer in this line represents a_i, j.

It is guaranteed that all the given numbers a_i, j are positive and do not exceed s. It is guaranteed that all a_i, j are distinct.

Output

In the first line print a single integer m(0 ≤ m ≤ s), representing the number of performed swaps.

In the next m lines print the description of these swap operations. In the i-th line print four space-separated integers x_i, y_i, p_i, q_i(1 ≤ x_i, p_i ≤ n; 1 ≤ y_i ≤ c_xi; 1 ≤ q_i ≤ c_pi). The printed numbers denote swapping the contents of cells a_xi, yi and a_pi, qi. Note that a swap operation can change the contents of distinct table cells. Print the swaps in the order, in which they should be executed.

Sample Input

Input

3 3 2 1 4 3 5 6 1 2

Output

2 1 1 2 2 2 1 3 1

Input

1 4 4 3 2 1

Output

2 1 1 1 4 1 2 1 3
题解：给N个数，接下来N个数，每个数代表i行的元素个数，要保证数列从左到右递增，从上到下递增；输出一种方案。交换次数，接下来每行输出交换的位置；
刚开始看错了，还以为最少交换次数，没敢写。。。
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = ;
int c[MAXN];
int a[MAXN][MAXN];
struct Node{
    int x, y;
    void init(int x = , int y = ){
        this->x = x;
        this->y = y;
    }
};
Node dt[MAXN * MAXN];
struct Nd{
    int x, y, nx, ny;
    void init(int x = , int y = , int nx = , int ny = ){
        this->x = x;
        this->y = y;
        this->nx = nx;
        this->ny = ny;
    }
    void print(){
        printf("%d %d %d %d\n", x, y, nx, ny);
    }
};
Nd ans[MAXN * MAXN];
Nd operator + (Node a, Node b){
    Nd c;
    c.x = a.x;
    c.y = a.y;
    c.nx = b.x;
    c.ny = b.y;
    return c;
}
int main(){
    int N;
    while(~scanf("%d", &N)){
        memset(a, , sizeof(a));
        memset(dt, , sizeof(dt));
        for(int i = ; i < N; i++)
            scanf("%d", c + i);
        for(int i = ; i < N; i++){
            for(int j = ; j < c[i]; j++){
                scanf("%d", &a[i][j]);
                dt[a[i][j]].init(i + , j + );
            }
        }
        int cur = , cnt = ;
        Node x;
        for(int i = ; i < N; i++){
            for(int j = ; j < c[i]; j++){
            //    printf("cur = %d a[i][j] = %d\n", cur, a[i][j]);
                if(cur != a[i][j]){
                    x.init(i + , j + );
                    ans[cnt] = x + dt[cur];
                    a[dt[cur].x - ][dt[cur].y - ] = a[i][j];
                    dt[a[i][j]].init(dt[cur].x, dt[cur].y);
                    dt[cur].init(i + , j + );
                    cnt++;
                }
                cur++;
            }
        }
        printf("%d\n", cnt);
        for(int i = ; i < cnt; i++){
            ans[i].print();
        }
    }
    return ;
}