Problem: Cirno gives AquaMoon a problem. There are m people numbered from 0 to m - 1. They are standing on a coordinate axis in points with positive integer coordinates. They are facing right (i.e. in the direction of the coordinate increase). At this moment everyone will start running with the constant speed in the direction of coordinate increasing. The initial coordinate of the i-th person on the line is x_i, and the speed of the i-th person is v_i. So the coordinate of the i-th person at the moment t will be x_i + t ⋅ v_i.

Cirno captured the coordinates of m people in k consecutive integer moments from 0 to k - 1. In every moment, the coordinates of m people were recorded in arbitrary order.

To make the problem more funny, Cirno modified one coordinate at the moment y (0 < y < k-1) to a different integer.

AquaMoon wants to find the moment y and the original coordinate p before the modification. Actually, she is not a programmer at all. So she wasn't able to solve it. Can you help her?

Input

This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.

The first line contains two integers m and k (5 ≤ m ≤ 1000, 7 ≤ k ≤ 1000) — the number of people and the number of recorded moments. 

The next k lines contain captured positions. i-th of these lines contains m integers between 1 and 10^6 (inclusive), representing positions captured by Cirno at the moment i-1.

The input is guaranteed to be valid (i.e. only one integer was modified to a different value according to the problem statement). Also, it is guaranteed, that 1 ≤ v_i ≤ 1000 for all 1 ≤ i ≤ m.

Hack format:

The first line should contain two integers m and k (5 ≤ m ≤ 1000, 7 ≤ k ≤ 1000) — the number of people and the number of moments. 

In the second line, there should be m integers x_0, x_1, ...,x_{m - 1} (1 ≤ x_i ≤ 10^6), where x_i is the initial coordinate of the i-th person.

In the third line, there should be m integers v_0, v_1, ...,v_{m - 1} (1 ≤ v_i ≤ 1000), where v_i is the speed of the i-th person. It should be true that x_i + (k-1) v_i ≤ 10^6 for each 0 ≤ i < m.

In the next k lines, each line should contain m integers. i-th line should contain m distinct integers p_0, p_1, …, p_{m-1} (0 ≤ p_j < m). The meaning of these numbers: j-th integer in the input in the i-th moment is the coordinate of the p_{j}-th person.

In the last line, there should be three integers y, i, c. Cirno modified the coordinate of the i-th person at the moment y to c (1 ≤ y ≤ k-2, 0 ≤ i ≤ m - 1, 1 ≤ c ≤ 10^6, c ≠ x_i + y ⋅ v_i).

Output

Print a single line with two integers y, p — the moment that contains the modified coordinate and the original coordinate.

Example

Input


5 7
6 9 9 6 9
10 7 10 8 10
11 11 11 10 8
12 12 12 12 9
14 13 12 10 13
11 14 16 14 14
12 15 18 15 15


Output


4 13

Note

In the first test the initial coordinates of people are 9, 6, 6, 9, 9 and their speeds are 1, 2, 1, 1, 1. So, it's easy to see, that at the moment 4 one coordinate was modified from 13 to 12.

This is the first test in the hack format:
    
    
      
    5 7  
    9 6 6 9 9  
    1 2 1 1 1  
    2 3 4 1 0  
    0 2 3 1 4  
    4 3 0 1 2  
    1 3 4 0 2  
    1 4 0 2 3  
    2 4 1 3 0  
    2 4 1 3 0  
    4 0 12  
    

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std ;
#define ll long long
const ll N = 2000 ;
ll mod = 998244353 ;
ll a[ N ][ N ] , //原数据
c[ N ]; // 隔行位置和
int main(){
    ll n , m ;
    cin >> n >> m ;
    for( int i = 0 ; i < m ; i ++ ){
        for( int j = 1 ; j <= n ; j ++ ){
            cin >> a[ i ][ j ] ;
            c[ i ] += a[ i ][ j ] ;
        }
    }
    ll x = ( c[m-1] - c[0] )/( m - 1 ) ,//确定等差数列公差
     y , ans1 , ans2 ;
    for( int i = 1 ; i < m ; i ++ ){
        if( (c[i]-c[0])!= x*i ){
            ans1 = i ; // 找到有误的一行
            y = c[ i ] - c[ 0 ] - x*i ; //找到误差值
            break ;
        }
    }
    for( int i = 1 ; i < m - 1 ; i ++ ){
        if( i-1 != ans1 && i != ans1 && i+1 != ans1 ){//找到正常的三行
            x = 0 ;
            for( int j = 1 ; j <= n ; j ++ ){
                x += a[i-1][j]*a[i-1][j]+a[i+1][j]*a[i+1][j]-a[i][j]*a[i][j]*2;
                //计算2Σvi2
            }
            break ;
        }
    }
    ll t = 0 , s = 0 ;
    for( int j = 1 ; j <= n ; j ++ ){
        s += a[ans1-1][j]*a[ans1-1][j] + a[ans1+1][j]*a[ans1+1][j];//sum2[y-1]+sum2[y+1]
        t += a[ans1][j]*a[ans1][j]*2 ;//sum2[y]
    }
    s -= x ;//s=2sum2[y]
    // y = 误差值
    for( int j = 1 ; j <= n ; j ++ ){
        ll temp = t - a[ans1][j]*a[ans1][j]*2 + (a[ans1][j]-y)*(a[ans1][j]-y)*2 ;
        //枚举误差点
        if( temp == s ){
            ans2 = a[ans1][j] -y ;//计算正确的值
            break ;
        }
    }
    cout << ans1 << " " << ans2 ;
    return 0 ;

}