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 <iostream>
#include<bits/stdc++.h>
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,P> PP;
vector<PP> v[50];
ll n,m,s,a,b,c,d,C[50],D[50],dp[50][2500],cnt;
priority_queue<PP,vector<PP>,greater<PP>> que;
int main(void){
    cin>>n>>m>>s;
    for(int i=0;i<n;i++)for(int j=0;j<2500;j++)dp[i][j]=1e18;
    for(int i=0;i<m;i++){
        cin>>a>>b>>c>>d;
        a--,b--;
        v[a].push_back({b,{c,d}});
        v[b].push_back({a,{c,d}});
    }
    for(int i=0;i<n;i++){
        cin>>C[i]>>D[i];
    }
    if(s>=2500)s=2499;
    que.push({0,{0,s}});
    dp[0][s]=0;
    while(que.size()){
        auto p=que.top();
        que.pop();
        
        if(dp[p.S.F][p.S.S]<p.F)continue;
        
        if(p.S.S+C[p.S.F]<2500){
            if(dp[p.S.F][p.S.S+C[p.S.F]]>p.F+D[p.S.F]){
                dp[p.S.F][p.S.S+C[p.S.F]]=p.F+D[p.S.F];
                que.push({p.F+D[p.S.F],{p.S.F,p.S.S+C[p.S.F]}});
            }
        }
        for(auto x:v[p.S.F]){
            if(x.S.F>p.S.S)continue;
            if(dp[x.F][p.S.S-x.S.F]>p.F+x.S.S){
                dp[x.F][p.S.S-x.S.F]=p.F+x.S.S;
                que.push({p.F+x.S.S,{x.F,p.S.S-x.S.F}});
            }
        }
        
    
    }
    for(int i=1;i<n;i++){
        ll ans=1e18;
        for(int j=0;j<2500;j++){
            ans=min(ans,dp[i][j]);
        }
        cout<<ans<<endl;
    }
}
