#include <bits/stdc++.h>

using namespace std;

int n,m,a[23][23],num[23],flag=1;

int check(int x,int y)

{

    for(int i=1;i<=n;i++)

    {

        int cnt=0;

        for(int j=1;j<=m;j++)

        {

            if(j==x)

            {

                if(a[i][j]!=y)cnt++;

            }

            else if(j==y)

            {

                if(a[i][j]!=x)cnt++;

            }

            else

            {

                if(a[i][j]!=j)cnt++;

            }

        }

        if(cnt>2)return 0;

    }

    return 1;

}

int main()

{

    scanf("%d%d",&n,&m);

    for(int i=1;i<=n;i++)

        for(int j=1;j<=m;j++)

            scanf("%d",&a[i][j]);

    for(int i=1;i<=n;i++)

    {

        num[i]=0;

        for(int j=1;j<=m;j++)

        {

            if(a[i][j]!=j)num[i]++;

        }

        if(num[i]>4)

        {

            cout<<"NO";

            return 0;

        }

        if(num[i]>2)flag=0;

    }

    if(flag)

    {

        cout<<"YES";

        return 0;

    }

    flag=0;

    for(int i=1;i<=m;i++)

    {

        for(int j=i+1;j<=m;j++)

        {

            if(check(i,j))flag=1;

        }

    }

    if(flag)cout<<"YES";

    else cout<<"NO";

    return 0;

}

/*

题意：

每行交换最多一次，最后还可以交换两列,问最后每行是否都能变成1,2,...m的排列;

思路：

枚举最后要交换哪两列,然后判断每行是否能在一次变换中变成列交换之前的状态,还有判断不进行列交换的时候;

*/

There are k sensors located in the rectangular room of size n × m meters. The i-th sensor is located at point (xi, yi). All sensors are located at distinct points strictly inside the rectangle.

Opposite corners of the room are located at points (0, 0) and (n, m). Walls of the room are parallel to coordinate axes.

At the moment 0, from the point (0, 0) the laser ray is released in the direction of point (1, 1). The ray travels with a speed of meters per second. Thus, the ray will reach the point (1, 1) in exactly one second after the start.

When the ray meets the wall it's reflected by the rule that the angle of incidence is equal to the angle of reflection. If the ray reaches any of the four corners, it immediately stops.

For each sensor you have to determine the first moment of time when the ray will pass through the point where this sensor is located. If the ray will never pass through this point, print  - 1 for such sensors.

The first line of the input contains three integers n, m and k (2 ≤ n, m ≤ 100 000, 1 ≤ k ≤ 100 000) — lengths of the room's walls and the number of sensors.

Each of the following k lines contains two integers xi and yi (1 ≤ xi ≤ n - 1, 1 ≤ yi ≤ m - 1) — coordinates of the sensors. It's guaranteed that no two sensors are located at the same point.

Print k integers. The i-th of them should be equal to the number of seconds when the ray first passes through the point where the i-th sensor is located, or  - 1 if this will never happen.