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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
const LL MX = 5052;

struct node{
	LL x,dis;
	bool operator < (const node &nd)const{
		return nd.dis < dis;
	}
}h,t;

LL ec = 0,hed[2000005] = {0},nxt[8000005],to[8000005],w[8000005];
void add_edge(LL f,LL t,LL cst){
	++ ec; to[ec] = t; w[ec] = cst; nxt[ec] = hed[f]; hed[f] = ec;
}

// dijk
priority_queue <node> q;
LL dis[6000005];
LL vis[6000005];
void dijkstra(LL s){
	memset(dis,0x3f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	t.x = s; t.dis = 0; dis[s] = 0; q.push(t);
	while(!q.empty()){
		h = q.top(); q.pop();
		LL v,u = h.x; if(vis[u]) continue;
		vis[u] = 1;
		for(LL i = hed[u];i;i = nxt[i]){
			LL v = to[i];
			if(dis[v] > dis[u] + w[i]){
				dis[v] = dis[u] + w[i];
				t.x = v; t.dis = dis[v];
				q.push(t);
			}
		}
	}
}
// dijk

LL get_id(LL x,LL y){
	return (x - 1) * (MX + 1) + y + 1;
}
LL ans[2000005];

int main(){
	LL n,m,s,u,v,a,b;
	cin >> n >> m >> s;
	s = min(s,MX);
	for(LL i = 1;i <= m;i ++){
		cin >> u >> v >> a >> b;
		for(LL j = a;j <= MX;j ++){
			add_edge(get_id(u,j),get_id(v,j - a),b);
			add_edge(get_id(v,j),get_id(u,j - a),b);
		}
	}
	for(LL i = 1;i <= n;i ++){
		cin >> a >> b;
		for(LL j = 0;j <= MX;j ++) add_edge(get_id(i,j),get_id(i,min(j + a,MX)),b);
	}
	dijkstra(get_id(1,s));
	for(LL i = 1;i <= n;i ++){
		ans[i] = dis[get_id(i,0)];
		for(LL j = 1;j <= MX;j ++) ans[i] = min(ans[i],dis[get_id(i,j)]);
	}
	for(LL i = 2;i <= n;i ++) cout << ans[i] << '\n';
	return 0;
}