#include #include #define rep(i,n) for(int i=0;i vi; typedef vector vl; typedef vector> vvi; typedef vector> vvl; typedef long double ld; typedef pair P; template ostream& operator<<(ostream& os, const static_modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const dynamic_modint& a) {os << a.val(); return os;} template istream& operator>>(istream& is, static_modint& a) {long long x; is >> x; a = x; return is;} template istream& operator>>(istream& is, dynamic_modint& a) {long long x; is >> x; a = x; return is;} template istream& operator>>(istream& is, vector& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template ostream& operator<<(ostream& os, const pair& p){os << p.first << ' ' << p.second; return os;} template ostream& operator<<(ostream& os, const vector& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const vector>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template ostream& operator<<(ostream& os, const set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const unordered_set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const atcoder::segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const atcoder::lazy_segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template void chmin(T& a, T b){a = min(a, b);} template void chmax(T& a, T b){a = max(a, b);} template T pow_mod(T A, T N, T MOD){ T res = 1 % MOD; A %= MOD; while(N){ if(N & 1) res = (res * A) % MOD; A = (A * A) % MOD; N >>= 1; } return res; } const long long e9 = 1000000000; long long solve(long long n, long long m, long long MOD){ long long g = gcd(MOD, n); if(g >= 2){ if(m % g == 0) return solve(n / g, m /g, MOD / g); else return -1LL; } long long phi = MOD; if(MOD % 2 == 0) phi /= 2; if(MOD % 5 == 0){ phi /= 5; phi *= 4; } long long n_inv = pow_mod(n, phi - 1, MOD); long long ans = (((-m % MOD) + MOD) * n_inv) % MOD; if(ans == 0) ans += MOD; return ans; } int solve(){ long long n, m; cin >> n >> m; cout << solve(n, m, e9) << "\n"; return 0; }; int main(){ int t; cin >> t; rep(_, t) solve(); return 0; }