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 ll long long
using namespace std;
const int MS=2510,MN=60;
const ll INF=1e18;
struct edge{
    int to,money;
    ll  time;
    edge(){}
    edge(int _to,int _money,ll _time)
    {to=_to,money=_money,time=_time;}
    bool operator <(edge tmp) const
    {return time>tmp.time;}
};
vector<edge> g[MN];
int n,m,s;
ll dp[MN][MS+10],ans[MN];
int main()
{
    scanf("%d%d%d",&n,&m,&s);
    for(int i=0,u,v,a,b;i<m;i++)
    {
        scanf("%d%d%d%d",&u,&v,&a,&b);
        g[u].emplace_back(v,a,(ll)b);
        g[v].emplace_back(u,a,(ll)b);
    }
    for(int i=1,c,d;i<=n;i++)
    {
        scanf("%d%d",&c,&d);
        g[i].emplace_back(i,-c,(ll)d);
    }
    for(int i=0;i<MN;i++)
        for(int j=0;j<MS+10;j++)
            dp[i][j]=INF;
    for(int i=0;i<MN;i++)
        ans[i]=INF;
    priority_queue<edge> q;
    q.emplace(1,min(MS,s),0);
    while(!q.empty())
    {
        edge tmp=q.top();q.pop();
        int u=tmp.to,money=tmp.money;ll time=tmp.time;
        if(time>=dp[u][min(MS,money)])
            continue;
        dp[u][min(MS,money)]=time;
        ans[u]=min(time,ans[u]);
        for(auto e: g[u])
            if(money-e.money>=0)
                q.emplace(e.to,money-e.money,time+e.time);
    }
    for(int i=2;i<=n;i++)
        printf("%lld\n",ans[i]);
    return 0;
}
