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<ll,ll> pii;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
const ll inf=1LL<<60;
ll n,V;
ll gg(pii p){
    return p.first+p.second*n;
}
struct edge{
	ll to, cost;
};
vector<vector<edge> > g;
vector<ll> d;
void gao(ll s){
    priority_queue<pii,vector<pii>,greater<pii> > q;
    d=vector<ll>(V,inf);
    d[s]=0;
    q.push(pii(0,s));
    while(!q.empty()){
        pii p=q.top();q.pop();
        ll v=p.second;
        if(d[v]<p.first)continue;
		for(int i=0;i<g[v].size();i++){
			edge e=g[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                q.push(pii(d[e.to],e.to));
            }
        }
    }
}
int main(){
	n=read();
    ll m=read(),s=read();
    s=min(s,5000LL);
    V=n*6001;
    g=vector<vector<edge> >(n*6001);
    for(int i=0;i<m;i++){
        ll u=read()-1,v=read()-1,a=read(),b=read();
        for(int j=a;j<6001;j++){
            g[gg({u,j})].push_back({gg({v,j-a}),b});
            g[gg({v,j})].push_back({gg({u,j-a}),b});
        }
    }
    for(int i=0;i<n;i++){
        ll c=read(),d=read();
        for(int j=0;j<6001-c;j++)
            g[gg({i,j})].push_back({gg({i,j+c}),d});
    }
    gao(gg({0,s}));
    for(int i=1;i<n;i++){
        ll ans=inf;
        for(int j=0;j<6001;j++)
            ans=min(ans,d[gg({i,j})]);
        cout << ans << endl;
    }
	return 0;
}