Peter studies the theory of relational databases. Table in the relational database consists of values that are arranged in rows and columns.

There are different normal forms that database may adhere to. Normal forms are designed to minimize the redundancy of data in the database. For example, a database table for a library might have a row for each book and columns for book name, book author, and author's email.

If the same author wrote several books, then this rePResentation is clearly redundant. To formally define this kind of redundancy Peter has introduced his own normal form. A table is in Peter's Normal Form (PNF) if and only if there is no pair of rows and a pair of columns such that the values in the corresponding columns are the same for both rows.

The above table is clearly not in PNF, since values for 2rd and 3rd columns repeat in 2nd and 3rd rows. However, if we introduce unique author identifier and split this table into two tables -- one containing book name and author id, and the other containing book id, author name, and author email, then both resulting tables will be in PNF.

Given a table your task is to figure out whether it is in PNF or not.

Input 

Input contains several datasets. The first line of each dataset contains two integer numbers n and m ( 1n10000, 1m10), the number of rows and columns in the table. The following n lines contain table rows. Each row has m column values separated by commas. Column values consist of ASCII characters from space (ASCII code 32) to tilde (ASCII code 126) with the exception of comma (ASCII code 44). Values are not empty and have no leading and trailing spaces. Each row has at most 80 characters (including separating commas).

Output 

For each dataset, if the table is in PNF write to the output file a single Word ``YES" (without quotes). If the table is not in PNF, then write three lines. On the first line write a single word ``NO" (without quotes). On the second line write two integer row numbers r₁ and r₂ ( 1r₁, r₂n, r₁r₂), on the third line write two integer column numbers c₁ and c₂ ( 1c₁, c₂m, c1c2), so that values in columns c₁and c₂ are the same in rows r₁ and r₂.

Sample Input 

3 3How to compete in ACM ICPC,	Peter,		peter@neerc.ifmo.ruHow to win ACM ICPC,		Michael,	michael@neerc.ifmo.ruNotes from ACM ICPC champion,	Michael,	michael@neerc.ifmo.ru2 31,Peter,peter@neerc.ifmo.ru2,Michael,michael@neerc.ifmo.ru
 
Sample Output 
 
NO2 32 3YES
題意：存在兩個不同行r1,r2和兩個不同列c1,c2。是否存在r1,r2和從c1,c2使得(r1,c1)和(r2,c1)相同。
分析：
可以直接寫一個四重循環枚舉出r1,r2,c1,c2。理論上是可以的，但實際上卻會TLE超時。
解決方法是只枚舉c1,c2,然后從上往下掃描各行。每次碰到一個新的行r,就把對應c1,c2的內容作為一個二元組存到一個map里，然后如果map的鍵值已經存在這個二元組，該二元組映射到的就是所要求的r1,而當前行就是r2。
細節問題：如何表示由c1,c2兩列組成的二元組？一種方法是直接用兩個字符串拼成一個長字符串（中間用一個其他地方不可能出現的字符分隔），但是速度比較慢，（因為在map中查找元素時需要進行字符串比較操作）。
更值得推薦的方法是在主循環之前先做一個預處理—給所有字符串分配一個編號，則整個數據庫中每個單元格都變成了整數，上述二元組就變成了兩個整數。
思路：1.首先將每一個表格里面的字符串用map進行編號處理2.一行一行的掃描，每兩列的編號作為一個二元組存入map中，若已經存在該編號則說明有滿足條件的，需要輸出。
#include <iostream>#include <string>#include <set>#include <vector>#include <map>using namespace std;const int ROW = 10000 + 10;const int COL = 10 + 5;int n,m;int s[ROW][COL];map<string, int> IDcache;struct node{    int x,y;    node(int x, int y):x(x),y(y) { }    bool Operator < (const node& r) const { return x<r.x || x==r.x&&y<r.y; }};map<node,int> data;int main(){    int n,m;    int ID=1;    cin>>n>>m;    string x;    for(int i=0;i<n;i++){        for(int j=0;j<m;j++){             cin>>x;             if(!IDcache.count(x)){                 IDcache[x]=ID;                 s[i][j]=ID;                 ID++;             }             else{                s[i][j]=IDcache[x];             }        }    }    int flag=0;        for(int c1=0;c1<m;c1++){        for(int c2=c1+1;c2<m;c2++){                data.clear();                for(int r=0;r<n;r++){                    int x = s[r][c1];                    int y = s[r][c2];                    node p(x,y);                    if(!data.count(p)){                        data[p]=r;                    }                    else                    {                        flag=1;                        cout<<"NO"<<endl;                        cout<<data[p]+1<<" "<<r+1<<endl<<c1+1<<" "<<c2+1<<endl;                    }                }        }    }    if(flag==0)    {      cout<<"YES";    }    cout<data++<<endl;        return 0;}