#include "bits/stdc++.h"
using namespace std;
#define int long long
#define double long double
#define INF 1000000000000000000
#define MOD 1000000007
int n;
int query(int x, int y) {
if(x <= 0 || x > n || y <= 0 || y > n) return 0;
cout << "examine " << x << " " << y << endl;
string s;
cin >> s;
return s == "true";
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int x0, y0, L, R, U, D;
string s;
cin >> n >> x0 >> y0;
// solve in horizontal dim
int l = 1, r = x0, m;
while(l < r) {
m = (l+r)/2;
if(query(m, y0)) r = m;
else l = m+1;
}
L = l;
l = x0+1; r = n+1;
while(l < r) {
m = (l+r)/2;
if(query(m, y0)) l = m+1;
else r = m;
}
R = l-1;
// cout << "L R " << L << " " << R << endl;
if(query((L+R)/2, y0)) {
// check if (not) xoxox
if(query( L-1+(R-L+1)*4/5, y0 )) { // x
int L1 = L, R1 = R;
if(query( L-1 + (R-L+1)*5, y0 )) R1 = L-1 + (R-L+1)*5;
else if(query( L-1 + (R-L+1)*3, y0 )) R1 = L-1 + (R-L+1)*3;
if(query( R+1 - (R-L+1)*5, y0 )) L1 = R+1 - (R-L+1)*5;
else if(query( R+1 - (R-L+1)*3, y0 )) L1 = R+1 - (R-L+1)*3;
L = L1; R = R1;
}
} else { // __xox__
if(query( L-1+ (R-L+1)*5/3, y0 )) R = L-1+ (R-L+1)*5/3;
else if(query( R+1 - (R-L+1)*5/3, y0 )) L = R+1 - (R-L+1)*5/3;
else {
int R1 = L-1 + (R-L+1)*4/3;
int L1 = R+1 - (R-L+1)*4/3;
L = L1; R = R1;
}
}
// cout << "final L R " << L << " "<< R << endl;
// solve in vertical dim
l = 1; r = y0;
while(l < r) {
m = (l+r)/2;
if(query(x0, m)) r = m;
else l = m+1;
}
D = l;
l = y0+1; r = n+1;
while(l < r) {
m = (l+r)/2;
if(query(x0, m)) l = m+1;
else r = m;
}
U = l-1;
// cout << "D U " << D << " " << U << endl;
if(query(x0, (D+U)/2)) {
// check if (not) xoxox
if(query( x0, D-1+(U-D+1)*4/5 )) { // x
int D1 = D, U1 = U;
if(query( x0, D-1 + (U-D+1)*5 )) U1 = D-1 + (U-D+1)*5;
else if(query( x0, D-1 + (U-D+1)*3 )) U1 = D-1 + (U-D+1)*3;
if(query( x0, U+1 - (U-D+1)*5 )) D1 = U+1 - (U-D+1)*5;
else if(query( x0, U+1 - (U-D+1)*3 )) D1 = U+1 - (U-D+1)*3;
D = D1; U = U1;
}
} else { // __xox__
if(query( x0, D-1+ (U-D+1)*5/3 )) U = D-1+ (U-D+1)*5/3;
else if(query( x0, U+1 - (U-D+1)*5/3 )) D = U+1 - (U-D+1)*5/3;
else {
int U1 = D-1 + (U-D+1)*4/3;
int D1 = U+1 - (U-D+1)*4/3;
D = D1; U = U1;
}
}
// cout << "final D U " << D << " "<< U << endl;
cout << "solution " << (L+R)/2 << " " << (U+D)/2 << endl;
}
