分析

• 由图可得：一开始放一个，由于边上的无需依靠对角线 $rebuild$ ，所以可以直接 $rebuild$ ，这样一列就好了。然后第二列随便放一个位置，其他所有都可以靠对角线 $rebuild$ 。剩余的同理。所以答案应该是行列最大值。

Think Twice, Code Once

#include <bits/stdc++.h>
#define il inline
#define get getchar
#define put putchar
#define is isdigit
#define int long long
#define dfor(i,a,b) for(int i=a;i<=b;++i)
#define dforr(i,a,b) for(int i=a;i>=b;--i)
#define dforn(i,a,b) for(int i=a;i<=b;++i,put(10))
#define mem(a,b) memset(a,b,sizeof a)
#define memc(a,b) memcpy(a,b,sizeof a)
#define pr 114514191981
#define gg(a) cout<<a,put(32)
#define INF 0x7fffffff
#define tt(x) cout<<x<<'\n'
#define endl '\n'
#define ls i<<1
#define rs i<<1|1
#define la(r) tr[r].ch[0]
#define ra(r) tr[r].ch[1]
#define lowbit(x) (x&-x)
#define ct cin.tie(nullptr),ios_base::sync_with_stdio(false)
using namespace std;
typedef unsigned int ull;
typedef pair<int, int> pii;
int read(void) {
    int x=0,f=1;char c=get();
    while(!is(c)) (f=c==45?-1:1),c=get();
    while(is(c)) x=(x<<1)+(x<<3)+(c^48),c=get();
    return x*f;
}
void write(int x) {
    if (x < 0) x = -x, put(45);
    if (x > 9) write(x / 10);
    put((x % 10) ^ 48);
}
#define writeln(a) write(a), put(10)
#define writesp(a) write(a), put(32)
#define writessp(a) put(32), write(a)
const int N = 1e5 + 10, M = 1e5 + 10, SN = 1e3 + 10, mod = 998244353;
signed main() {
    int T = read();
    while (T--) {
        int n = read(), m = read();
        writeln(max(n, m));
    }
    return 0;
}