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;
const int mxp=2500;

int n,m,s;
vector<pair<int,int>> g[50*(mxp+1)+5];
ll d[50*(mxp+1)+5];
int give[55],cost[55];

inline int id(const int &ver,const int &poss){
    return (ver-1)*(mxp+1)+poss;
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n>>m>>s;
    s=min(s,mxp);
    for(int i=1;i<=m;i++){
        int u,v,a,b;
        cin>>u>>v>>a>>b;
        for(int i=a;i<=mxp;i++){
            g[id(u,i)].emplace_back(id(v,i-a),b);
            g[id(v,i)].emplace_back(id(u,i-a),b);
        }
    }
    for(int i=1;i<=n;i++){
        cin>>give[i]>>cost[i];
        give[i]=min(give[i],mxp);
        for(int j=0;j+give[i]<=mxp;j++){
            g[id(i,j)].emplace_back(id(i,j+give[i]),cost[i]);
        }
    }
    priority_queue<pair<ll,int>,vector<pair<ll,int>>,greater<pair<ll,int>>> pq;
    pq.push(make_pair(0,id(1,s)));
    memset(d,0x3f,sizeof(d));
    d[id(1,s)]=0;
    while(!pq.empty()){
        ll cd=pq.top().first;
        int x=pq.top().second;
        pq.pop();
        if(d[x]<cd)continue;
        for(pair<int,int> &p:g[x]){
            if(d[p.first]>d[x]+p.second){
                d[p.first]=d[x]+p.second;
                pq.push(make_pair(d[p.first],p.first));
            }
        }
    }
    for(int i=2;i<=n;i++){
        ll ans=1e18;
        for(int j=0;j<=mxp;j++){
            ans=min(ans,d[id(i,j)]);
        }
        cout<<ans<<endl;
    }

    return 0;
}