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 int long long 
const int INF=1e16;
const int MAXN=69,MAXS=25e2+69;
struct vl{
	int x,y,z;
	bool operator<(const vl& oth)const{
		if(x==oth.x&&y==oth.y)return z<oth.z;
		if(x==oth.x)return y<oth.y;
		return x<oth.x;
	}
};
int n,m,S;
vector<vl>g[MAXN];
vl shop[MAXN];
int kc[MAXN][MAXS];
void ditcha(int u){
	for(int i=1;i<=n;i++){
		for(int j=0;j<MAXS;j++)kc[i][j]=INF;
	}
	set<vl>s;
	int i=0;
	while(1){
		int c=S+i*shop[u].x,br=0;
		if(c>=MAXS-1)br=1,c=MAXS-1;
		kc[u][c]=i*shop[u].y;
		s.insert({kc[u][c],u,c});
		i++;
		if(br)break;
	}
	while(!s.empty()){
		int u=(*s.begin()).y;
		int c=(*s.begin()).z;
		s.erase(s.begin());
		for(vl i:g[u]){
			int v=i.x,a=i.y,b=i.z;
			if(c-a<0)continue;
			int j=0;
			while(1){
				int d=c-a+j*shop[v].x,br=0;
				if(d>=MAXS-1)br=1,d=MAXS-1;
				if(kc[v][d]>kc[u][c]+b+j*shop[v].y){
					if(kc[v][d]!=INF)s.erase({kc[v][d],v,d});
					kc[v][d]=kc[u][c]+b+j*shop[v].y;
					s.insert({kc[v][d],v,d});
				}
				j++;
				if(br)break;
			}
		}
	}
}
signed main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m>>S;
	for(int i=1,u,v,a,b;i<=m;i++){
		cin>>u>>v>>a>>b;
		g[u].push_back({v,a,b});
		g[v].push_back({u,a,b});
	}
	for(int i=1;i<=n;i++)cin>>shop[i].x>>shop[i].y;
	ditcha(1);
	for(int v=2;v<=n;v++){
		int res=INF;
		for(int i=0;i<MAXS;i++)res=min(res,kc[v][i]);
		cout<<res<<'\n';
	}
}