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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<ll,int> P;
typedef pair<P,int> PI;

int n,m,s;
ll dp[51][10001];
vector<PI> G[51];
int c[51],d[51];

int main(void){
	scanf("%d%d%d",&n,&m,&s);
	for(int i=0;i<m;i++){
		int u,v,a,b;
		scanf("%d%d%d%d",&u,&v,&a,&b);
		u--;
		v--;
		G[u].push_back(PI(P(v,a),b));
		G[v].push_back(PI(P(u,a),b));
	}
	for(int i=0;i<n;i++){
		scanf("%d%d",&c[i],&d[i]);
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<=10000;j++){
			dp[i][j]=1LL<<50;
		}
	}
	dp[0][min(10000,s)]=0;
	priority_queue<PI,vector<PI>,greater<PI> > que;
	que.push(PI(P(0,0),min(10000,s)));
	while(que.size()){
		PI p=que.top();
		que.pop();
		int v=p.first.second;
		int nc=min(10000,p.second+c[v]);
		if(dp[v][nc]>dp[v][p.second]+d[v]){
			dp[v][nc]=dp[v][p.second]+d[v];
			que.push(PI(P(dp[v][nc],v),nc));
		}
		for(int i=0;i<G[v].size();i++){
			int nv=G[v][i].first.first;
			if(p.second>=G[v][i].first.second){
				int nc=p.second-G[v][i].first.second;
				ll nd=dp[v][p.second]+G[v][i].second;
				if(dp[nv][nc]>nd){
					dp[nv][nc]=nd;
					que.push(PI(P(nd,nv),nc));
				}
			}
		}
	}
	for(int i=1;i<n;i++){
		ll ans=dp[i][0];
		for(int j=0;j<=10000;j++){
			ans=min(ans,dp[i][j]);
		}
		printf("%lld\n",ans);
	}
	return 0;
}
