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.

Examples

Input

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


Output

2
14


Input

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


Output

5
5
7


Input

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


Output

1
9003
14606
16510
16576


Input

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


Output

1
3
5


Input

2 1 0
1 2 1 1
1 1000000000
1 1


Output

1000000001
#include<bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define print(A,x,n) rep(i,0,n){cout<<(i ? " ":"")<<A[i]x;}cout<<endl;
#define pprint(A,y,m,n) rep(j,0,m){print(A[j],y,n);}
const long mod=1e9+7;
const int size=3e5;
const long inf=1e14;
int N,M; int S;
vector<pair<pair<int,int>,long>> g[50][2500];
long dist[50][2500];
void dijkstra(int a){// O(NlogN+M)
	priority_queue<pair<long,pair<int,int>>> pq;
	rep(i,0,N)rep(j,0,2500){
		if(i == 0 && j == S){ dist[0][S] = 0; pq.push({0, {0, S}}); }
		else{ dist[i][j] = inf; pq.push({-inf, {i, j}}); }
	}
	while(!pq.empty()){
		int v = pq.top().second.first;
		int t = pq.top().second.second;
		pq.pop();
		for(pair<pair<int,int>,long> p:g[v][t]){
            int u = p.first.first;
            int k = p.first.second;
			long temp = dist[v][t] + p.second;
			if(dist[u][k] > temp){
				dist[u][k] = temp;
				pq.push({-dist[u][k], {u, k}});
			}
		}
	}return;
}
int main(){
	cin>>N>>M>>S; S = min(S, 2499);
	int a; long b;
	rep(i,0,M){
		int u,v;
		cin>>u>>v>>a>>b; u--; v--;
		rep(t,a,2500){
			g[u][t].push_back({{v, t-a}, b});
			g[v][t].push_back({{u, t-a}, b});
		}
	}
	int c; long d;
	rep(i,0,N){
        cin>>c>>d;
		rep(t,0,2500-c){
			g[i][t].push_back({{i, t+c}, d});
		}
	}
	dijkstra(0);
	long ans = 0;
	for(long i = 1; i < N; i++){
		ans = inf;
		for(long j = 0; j < 2500; j++) ans = min(ans, dist[i][j]);
		cout<<ans<<endl;
	}//pprint(dist,,N,10);
}