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 mp make_pair
typedef long long ll;
typedef pair<int,int>pii;
typedef pair<int,pii>piii;
typedef pair<ll,pii>plii;
const int Max = 3000;
const int N = 50;
const ll inf = 1e18;
ll dist[N+5][Max+10];
vector<piii>g[N+5];
int c[N+5],d[N+5];

int main()
{
	int n,m,s;
	scanf("%d%d%d",&n,&m,&s);
	s = min(s,Max);
	for(int i=1;i<=m;i++){
		int u,v,a,b;
		scanf("%d%d%d%d",&u,&v,&a,&b);
		g[u].push_back(mp(v,mp(a,b)));
		g[v].push_back(mp(u,mp(a,b)));
	}
	for(int i=1;i<=n;i++){
		scanf("%d%d",c+i,d+i);
	}
	for(int i=1;i<=N;i++){
		for(int j=0;j<=Max;j++){
			dist[i][j] = inf;
		}
	}
	priority_queue<plii,vector<plii>,greater<plii> >pq;
	pq.push(mp(0,mp(1,s)));
	dist[1][s] = 0;
	while(!pq.empty())
	{
		plii cur = pq.top();
		pq.pop();
		int v = cur.second.first;
		int sil = cur.second.second;
		ll t = cur.first;
		if(dist[v][sil]!=t){
			continue;
		}
		if(dist[v][min(sil+c[v],Max)]>t+d[v]){
			dist[v][min(sil+c[v],Max)] = t+d[v];
			pq.push(mp(t+d[v],mp(v,min(sil+c[v],Max))));
		}
		for(auto it:g[v])
		{
			int u = it.first;
			int sill = it.second.first;
			int tm = it.second.second;
			if(sil-sill>=0){
				if(dist[u][sil-sill]>t+tm){
					dist[u][sil-sill] = t+tm;
					pq.push(mp(t+tm,mp(u,sil-sill)));
				}
			}
		}
	}
	for(int i=2;i<=n;i++){
		ll ans = inf;
		for(int j=0;j<=Max;j++){
			ans = min(ans,dist[i][j]);
		}
		printf("%lld\n",ans);
	}
}
