#include <bits/stdc++.h> 
using namespace std;
 
typedef signed long long ll;
typedef pair<ll, ll> pll;
const ll mod = 1e9 + 7;
map<pll, int> mp;
pll Arr[20];
int n;
 
void solve(ll a, ll b, int i){
	if(i == n){
		ll mn = min(a, b), mx = max(a, b);
		mp[pll(mn, mx)]++;
	}else{
		ll sz = Arr[i].second, p = Arr[i].first, _a, _b;
		_a = ((a % mod) * ((ll)pow(p, sz)) % mod) % mod;
 
		for(ll j = 0; j <= sz; j++){
			_b = ((b % mod) * ((ll)pow(p, j)) % mod) % mod;
			solve(_a, _b, i + 1);
		}
 
		_b = ((b % mod) * ((ll)pow(p, sz)) % mod) % mod;
		for(ll j = 0; j <= sz; j++){
			_a = ((a % mod) * ((ll)pow(p, j)) % mod) % mod;
 
			solve(_a, _b, i + 1);
		}
	}
}
 
void solve(int i){
	ll sum = 0;
	solve(1LL, 1LL, 0);
	map<pll, int>:: iterator it;
	for(it = mp.begin(); it != mp.end(); it++){
		pll pr = it->first;
		sum = (sum + (pr.first + pr.second) % mod) % mod;
	}
	printf("Case %d: %lld\n", i, sum);
}
 
int main(){
	int t;
	ll p, cnt;
	scanf("%d", &t);
	for(int i = 1; i <= t; i++){
		scanf("%d", &n);
		for(int j = 0; j < n; j++){
			scanf("%lld %lld", &p, &cnt);
			Arr[j] = pll(p, cnt);
		}
 
		mp.clear();
		solve(i);
	}
	return 0;
}

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