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
//This program is written by Brian Peng.
#pragma GCC optimize("Ofast","inline","no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define Rd(a) (a=rd())
#define Gc(a) (a=getchar())
#define Pc(a) putchar(a)
int rd(){
	int x;char c(getchar());bool k;
	while(!isdigit(c)&&c^'-')if(Gc(c)==EOF)exit(0);
	c^'-'?(k=1,x=c&15):k=x=0;
	while(isdigit(Gc(c)))x=x*10+(c&15);
	return k?x:-x;
}
void wr(int a){
	if(a<0)Pc('-'),a=-a;
	if(a<=9)Pc(a|'0');
	else wr(a/10),Pc((a%10)|'0');
}
signed const INF(0x3f3f3f3f),NINF(0xc3c3c3c3);
long long const LINF(0x3f3f3f3f3f3f3f3fLL),LNINF(0xc3c3c3c3c3c3c3c3LL);
#define Ps Pc(' ')
#define Pe Pc('\n')
#define Frn0(i,a,b) for(int i(a);i<(b);++i)
#define Frn1(i,a,b) for(int i(a);i<=(b);++i)
#define Frn_(i,a,b) for(int i(a);i>=(b);--i)
#define Mst(a,b) memset(a,b,sizeof(a))
#define File(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
#define N (60)
int n,m,s,ans[N]{0,-1},u,v,a,b,c[N],d[N],mx[N];
struct E{int v,a,b;};
vector<E>e[N];
struct Q{int u,m,d;bool operator<(Q b)const{return d>b.d;}}t;
priority_queue<Q>q;
signed main(){
	*ans=Rd(n)-1,Rd(m),q.push({1,rd(),0}),Mst(mx,-1);
	while(m--)Rd(u),Rd(v),s+=Rd(a),Rd(b),e[u].push_back({v,a,b}),e[v].push_back({u,a,b});
	Frn1(i,1,n)Rd(c[i]),Rd(d[i]);
	while(!q.empty()){
		t=q.top(),q.pop(),t.m=min(t.m,s);
		if(mx[t.u]>=t.m)continue;mx[t.u]=t.m;
		if(!ans[t.u]){ans[t.u]=t.d;if(!--*ans)break;}
		q.push({t.u,t.m+c[t.u],t.d+d[t.u]});
		for(E i:e[t.u])if(i.a<=t.m)q.push({i.v,t.m-i.a,t.d+i.b});
	}
	Frn1(i,2,n)wr(ans[i]),Pe;
	exit(0);
}
