分三类
1 1: 一个就好了
3 3:特殊讨论下
n≥4 或 m≥4 : 第一行奇序号的数放前面，偶序号的数放后面，第二行奇序号的数放前面，偶序号的数放后面，第二行依次类推
有点难写，真的菜

#include<iostream>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<cmath>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5 + 5;

#define MS(x,y) memset(x,y,sizeof(x))

#define MP(x, y) make_pair(x, y)

const int INF = 0x3f3f3f3f;

map<pair<int, int>, int> mp;

int vc[N];

int main() {

    int n, m;

    while(~scanf("%d %d", &n, &m)) {

        mp.clear();

        for(int i = 0; i < n; ++i) {

            for(int j = 0; j < m; ++j) {

                mp[MP(i, j)] = i*m + j;

            }

        }

        if(n == 1 && m == 1) {

            printf("YES\n1\n");

        }else if(n == 3 && m == 3) {

            printf("YES\n");

            printf("6 1 8\n7 5 3\n2 9 4\n");

        }else if(n >= 4) {

            printf("YES\n");

            for(int i = 0; i < m; ++i) {

                int cnt = 0;

                for(int j = 0; j < n; ++j) {

                    vc[cnt ++] = mp[MP(j, i)];

                }

                if(n == 4) {

                    if(i & 1) { mp[MP(0, i)] = vc[2]; mp[MP(1, i)] = vc[0]; mp[MP(2, i)] = vc[3]; mp[MP(3, i)] = vc[1]; }

                    else      { mp[MP(0, i)] = vc[1]; mp[MP(1, i)] = vc[3]; mp[MP(2, i)] = vc[0]; mp[MP(3, i)] = vc[2]; }

                    continue;

                }

                int tt = 0;

                if(i & 1) {

                    for(int j = 0; j < cnt; j += 2) mp[MP(tt++, i)] = vc[j];

                    for(int j = 1; j < cnt; j += 2) mp[MP(tt++, i)] = vc[j];

                }else {

                    for(int j = 1; j < cnt; j += 2) mp[MP(tt++, i)] = vc[j];

                    for(int j = 0; j < cnt; j += 2) mp[MP(tt++, i)] = vc[j];

                }

            }

            for(int i = 0; i < n; ++i) {

                for(int j = 0; j < m; ++j) printf("%d ", mp[MP(i, j)] + 1);

                printf("\n");

            }

        }else if(m >= 4) {

            printf("YES\n");

            for(int i = 0; i < n; ++i) {

                int cnt = 0;

                for(int j = 0; j < m; ++j) {

                    vc[cnt ++] = mp[MP(i, j)];

                }

                if(m == 4) {

                    if(i & 1) { mp[MP(i, 0)] = vc[2]; mp[MP(i, 1)] = vc[0]; mp[MP(i, 2)] = vc[3]; mp[MP(i, 3)] = vc[1]; }

                    else      { mp[MP(i, 0)] = vc[1]; mp[MP(i, 1)] = vc[3]; mp[MP(i, 2)] = vc[0]; mp[MP(i, 3)] = vc[2]; }

                    continue;

                }

                int tt = 0;

                if(i & 1) {

                    for(int j = 0; j < cnt; j += 2) mp[MP(i, tt++)] = vc[j];

                    for(int j = 1; j < cnt; j += 2) mp[MP(i, tt++)] = vc[j];

                }

                else {

                    for(int j = 1; j < cnt; j += 2) mp[MP(i, tt++)] = vc[j];

                    for(int j = 0; j < cnt; j += 2) mp[MP(i, tt++)] = vc[j];

                }

            }

            for(int i = 0; i < n; ++i) {

                for(int j = 0; j < m; ++j) printf("%d ", mp[MP(i, j)] + 1);

                printf("\n");

            }

        }else printf("NO\n");

    }

    return 0;

}