This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "game.h"
#include <bits/stdc++.h>
#define endl "\n"
#define F first
#define S second
#define pb push_back
#define all(x) x.begin() , x.end()
#define mm(dp , val) memset (dp , val , sizeof dp)
#define mid ((r+l)>>1)
#define lx (n<<1)
#define rx ((n<<1)|1)
#define low (i&(-i))
#define lb lower_bound
#define ub upper_bound
#define no void (cout << "NO" << endl)
#define one void (cout << "-1" << endl)
#define zer void (cout << "0" << endl)
#define yes void (cout << "YES" << endl)
typedef long long ll;
using namespace std;
const int logn = 26 ;
const int N = 2e3+5;
ll dx[] = {0 , 0 , 1 , -1};
ll dy[] = {1 , -1 , 0 , 0};
long long gcd2(long long X, long long Y)
{
    if (!X || !Y)
        return max(X , Y) ;
    long long tmp;
    while (X != Y && Y != 0) {
        tmp = X;
        X = Y;
        Y = tmp % Y;
    }
    return X;
}
ll n , m , a[N][N] ;
struct seg_tree
{
    ll gcd[N] ;
    void upd (ll n , ll l , ll r , ll i , ll v)
    {
        if (l == r)
        {
            gcd[n] = v ;
            return ;
        }
        if (i <= mid)
            upd(lx , l , mid , i , v) ;
        else
            upd(rx , mid+1 , r , i , v) ;
        gcd[n] = gcd2(gcd[lx] , gcd[rx]) ;
    }
    ll qry (ll n , ll l , ll r , ll x , ll y)
    {
        if (l > y || r < x)
            return 0  ;
        if (x <= l && r<=y)
            return gcd[n] ;
        if (y <= mid)
            return qry(lx , l , mid , x , y) ;
        if (x > mid)
            return qry(rx , mid+1 , r , x , y) ;
        ll le = qry(lx , l , mid , x , y) ;
        ll ri = qry(rx , mid+1 , r , x , y) ;
        return gcd2(le , ri) ;
    }
} tree[N] , tree2[N];
ll get (ll x1 , ll y1 , ll x2 , ll y2)
{
    ll ret = 0 ;
    if (x1 - x2 <= y1 - y2)
    {
        for (int i = x1 ; i<=x2 ; i++)
            ret = gcd2(ret , tree[i].qry (1 , 0 , m-1 , y1 , y2));
    }
    else
    {
        for (int i = y1 ; i<= y2 ; i++)
            ret = gcd2 (ret , tree2[i].qry(1 , 0 , n-1 , x1 , x2));
    }
    return ret ;
}
void init(int R, int C) {
   n = R , m = C ;
}
void update(int P, int Q, long long K)
{
    tree[P].upd(1 , 0 , m-1 , Q , K) ;
    tree2[Q].upd(1 , 0 , n-1 , P , K) ;
}
long long calculate(int P, int Q, int U, int V) {
    return get(P , Q , U , V) ;
}
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