Problem:
Score : 500 points

Problem Statement

There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes.
You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin.
The transaction takes D_i minutes for each gold coin you give.
You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq 100

0 \leq S \leq 10^9

1 \leq A_i \leq 50

1 \leq B_i,C_i,D_i \leq 10^9

1 \leq U_i < V_i \leq N

There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).

Each city t=2,...,N can be reached from City 1 with some number of railroads.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N

Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Sample Input 1

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5

Sample Output 1

2
14

The railway network in this input is shown in the figure below.

In this figure, each city is labeled as follows:

The first line: the ID number i of the city (i for City i)

The second line: C_i / D_i

Similarly, each railroad is labeled as follows:

The first line: the ID number i of the railroad (i for the i-th railroad in input)

The second line: A_i / B_i

You can travel from City 1 to City 2 in 2 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

You can travel from City 1 to City 3 in 14 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 6 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Sample Input 2

4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4

Sample Output 2

5
5
7

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 4 in 7 minutes, as follows:

At the exchange counter in City 1, exchange 2 gold coins for 6 silver coins in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Use the 4-th railroad to move from City 3 to City 4 in 1 minutes.

Sample Input 3

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1

Sample Output 3

1
9003
14606
16510
16576

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 6 in 16576 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 9000 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

At the exchange counter in City 3, exchange 8 gold coins for 8 silver coins in 5600 minutes.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Use the 3-rd railroad to move from City 2 to City 4 in 1 minute.

At the exchange counter in City 4, exchange 19 gold coins for 19 silver coins in 1900 minutes.

Use the 3-rd railroad to move from City 4 to City 2 in 1 minute.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

Use the 4-th railroad to move from City 3 to City 5 in 1 minute.

At the exchange counter in City 5, exchange 63 gold coins for 63 silver coins in 63 minutes.

Use the 4-th railroad to move from City 5 to City 3 in 1 minute.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 5-th railroad to move from City 1 to City 6 in 1 minute.

Sample Input 4

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7

Sample Output 4

1
3
5

The railway network in this input is shown in the figure below:

Sample Input 5

2 1 0
1 2 1 1
1 1000000000
1 1

Sample Output 5

1000000001

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 2 in 1000000001 minutes, as follows:

At the exchange counter in City 1, exchange 1 gold coin for 1 silver coin in 1000000000 minutes.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define ld long double
#define pb push_back
 
#define sec second
#define fir first
#define mo  998244353
#define inf 1e18
#define rep(i, s, n) for (ll i = s; i < n; i = i + 1)
#define rrep(i,s,n) for(ll i=s;i>=n;i--)
#define adiii\                               
  ios_base::sync_with_stdio(false);\                                      
  cin.tie(NULL);\
  cout.tie(NULL);									
 
ll mod(ll n) { return (n % (ll)mo + (ll)mo)%(ll)mo;}
ll gcd(ll a,ll b) {if (b == 0) return a;return gcd(b, a % b);}

int main(){
	adiii
	ll n,m,s;
	cin>>n>>m>>s;
	s = min(s,(ll)2501);
	vector<pair<ll,pair<ll,ll>>> adj[n+1];
	rep(i,0,m){
		ll u,v,a,b;
		cin>>u>>v>>a>>b;
		adj[u].pb({v,{a,b}});
		adj[v].pb({u,{a,b}});
	}
    vector <ll> c(n+1),d(n+1);
    rep(i,1,n+1) cin>>c[i]>>d[i];

    
    ll time[n+1][2501];
    rep(i,1,n+1) rep(j,0,2501) time[i][j] = inf;
    time[1][s] = 0;
    using P = pair<ll,ll>;
    priority_queue<pair<ll,P> , vector<pair<ll,P>>, greater<pair<ll,P>>>pq;
    pq.push({0,{1,s}});
	while(!pq.empty()){
		ll u = pq.top().second.first;
		ll cs = pq.top().second.second;
		ll du = pq.top().first;
		//cout<<u<<" "<<cs<<endl;
		pq.pop();
		if(time[u][cs] != du) continue;
		if(time[u][cs] + d[u] < time[u][min(cs+c[u],(ll)2500)]){
			time[u][min(cs + c[u],(ll)2500)] = time[u][cs] + d[u];
			pq.push({time[u][cs] + d[u],{u,min(cs + c[u],(ll)2500)}});
		}
		for(auto vp:adj[u]){
		    ll v = vp.first;
		    ll tn =  vp.second.second;
		    ll cr = vp.second.first;
		    if(cr > cs) continue;
		    if(time[v][cs - cr] > time[u][cs] + tn){
				//cout<<v<<" "<<cs - cr<<endl;
				time[v][cs - cr] = time[u][cs] + tn;
				pq.push({time[v][cs-cr],{v,cs-cr}});
			}
		}
	}
	rep(i,2,n+1){
		ll ans = inf;
		rep(j,0,2501){
		  ans = min(ans,time[i][j]);
		}
		cout<<ans<<endl;
	}
    
    
}