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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

template<class T> struct edge{
	int to;
	T wt1,wt2;
	edge(int to,const T& wt1,const T& wt2):to(to),wt1(wt1),wt2(wt2){}
};
template<class T> using weighted_graph=vector<vector<edge<T>>>;

const long long INF=1LL<<61;

int main(){
	int n,m,k0; scanf("%d%d%d",&n,&m,&k0);
	k0=min(k0,2500);
	weighted_graph<lint> G(n);
	rep(i,m){
		int u,v,c1,c2; scanf("%d%d%d%d",&u,&v,&c1,&c2); u--; v--;
		G[u].emplace_back(v,c1,c2);
		G[v].emplace_back(u,c1,c2);
	}
	vector<int> c1(n),c2(n);
	rep(u,n) scanf("%d%d",&c1[u],&c2[u]);

	lint d[50][2501];
	rep(u,n) rep(k,2501) d[u][k]=INF;
	d[0][k0]=0;

	priority_queue<pair<lint,pair<int,int>>> Q;
	Q.emplace(0,make_pair(0,k0));
	while(!Q.empty()){
		lint d0=-Q.top().first;
		int u,k; tie(u,k)=Q.top().second; Q.pop();

		if(d0>d[u][k]) continue;

		// 換金
		int k2=min(k+c1[u],2500);
		lint d2=d0+c2[u];
		if(d2<d[u][k2]){
			d[u][k2]=d2;
			Q.emplace(-d2,make_pair(u,k2));
		}

		// 移動
		for(auto e:G[u]) if(e.wt1<=k) {
			int v=e.to;
			k2=k-e.wt1;
			d2=d0+e.wt2;
			if(d2<d[v][k2]){
				d[v][k2]=d2;
				Q.emplace(-d2,make_pair(v,k2));
			}
		}
	}

	for(int u=1;u<n;u++) printf("%lld\n",*min_element(d[u],d[u]+2501));

	return 0;
}
