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;

typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,int> PLI;
const int MN=51;
const int MM=101;
const int MA=51;
const int MC=2*MN*MA;
const int MV=MN*MC+5;
const LL INF=0x3f3f3f3f3f3f3f3f;
// sp at most go through less tahn 2n node 2n-1 edges?
// V = 50*2*50*50=250 000
// E = 100*2*50*50=500 000

vector<PII> g[MV];

int n,m,s;

inline int id(int u,int val)
{
	return u*(MC)+val;
}

LL dis[MV];
int vis[MV];
priority_queue<PLI,vector<PLI>,greater<PLI>> qu; 
void dij(int st,int v)
{
	memset(dis,INF,sizeof(dis));
	qu.push({0,id(st,v)});
	dis[id(st,v)]=0;
	while(!qu.empty())
	{
		PLI p=qu.top(); qu.pop();
		int u=p.second;
		LL d=dis[u];
		if(vis[u]) continue;
		vis[u]=1;
		for(auto e:g[u])
		{
			int v=e.first;
			LL w=e.second;
			if(vis[v]) continue;
			if(dis[v]>d+w)
			{
				dis[v]=d+w;
				qu.push({dis[v],v});
			}
		}
	}
}

int main()
{
	scanf("%d%d%d",&n,&m,&s);
	int u,v,a,b;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d%d%d",&u,&v,&a,&b);
		for(int i=0;i+a<MC;i++)
		{
			g[id(u,i+a)].push_back({id(v,i),b});
			g[id(v,i+a)].push_back({id(u,i),b});
		}
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a,&b);
		for(int j=0;j+a<MC;j++)
			g[id(i,j)].push_back({id(i,j+a),b});
	}
	if(s>=MC) s=MC-1;
	dij(1,s);
	for(int i=2;i<=n;i++)
	{
		LL mi=INF;
		for(int j=0;j<MC;j++)
			mi=min(mi,dis[id(i,j)]);
		printf("%lld\n",mi);
	}
	return 0;
}