#include <iostream>
#include <vector>
using namespace std;

// 归并排序并计算逆序数
int mergeAndCount(vector<int>& nums, int left, int mid, int right) {
    vector<int> temp(right - left + 1);
    int i = left, j = mid + 1, k = 0;
    int inversion = 0;

    while (i <= mid && j <= right) {
        if (nums[i] <= nums[j]) {
            temp[k++] = nums[i++];
        } else {
            temp[k++] = nums[j++];
            inversion += (mid - i + 1);  // 统计逆序数
        }
    }

    while (i <= mid) temp[k++] = nums[i++];
    while (j <= right) temp[k++] = nums[j++];

    for (int p = 0; p < k; p++) {
        nums[left + p] = temp[p];
    }

    return inversion;
}

int countInversions(vector<int>& nums, int left, int right) {
    int inversion = 0;
    if (left < right) {
        int mid = left + (right - left) / 2;
        inversion += countInversions(nums, left, mid);
        inversion += countInversions(nums, mid + 1, right);
        inversion += mergeAndCount(nums, left, mid, right);
    }
    return inversion;
}

int main() {
    int m, n;
    cin >> m >> n;
    vector<vector<int>> grid(m, vector<int>(n));
    int zero_row = -1;

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            cin >> grid[i][j];
            if (grid[i][j] == 0) {
                zero_row = i;
            }
        }
    }

    // 展开为一维数组并去除0
    vector<int> nums;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid[i][j] != 0) {
                nums.push_back(grid[i][j]);
            }
        }
    }

    // 计算逆序数
    int inversion = countInversions(nums, 0, nums.size() - 1);

    // 计算空格所在行从下数的行号
    int row_from_bottom = m - zero_row;

    // 判断条件
    if (n % 2 == 1) {
        // 奇数列，仅看逆序数是否为偶数
        cout << (inversion % 2 == 0 ? "YES" : "NO") << endl;
    } else {
        // 偶数列，逆序数+空格行数的奇偶性是否为奇数
        int total = inversion + row_from_bottom;
        cout << (total % 2 == 1 ? "YES" : "NO") << endl;
    }

    return 0;
}