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;

int main()
{
    const ll M=5000;
    int n,m;
    ll s;
    scanf("%d%d%lld",&n,&m,&s);
    vector<array<ll,3>> v[n+1];
    for(int i=0;i<m;i++)
    {
        int a,b;
        ll x,y;
        scanf("%d%d%lld%lld",&a,&b,&x,&y);
        v[a].push_back({b,x,y});
        v[b].push_back({a,x,y});
    }
    vector<ll> c(n+1,0);
    vector<ll> d(n+1,0);
    for(int i=1;i<=n;i++) scanf("%lld%lld",&c[i],&d[i]);
    vector<vector<ll>> res(n+1,vector<ll>(M+1,(1ll<<62)));
    priority_queue<array<ll,3>,vector<array<ll,3>>,greater<array<ll,3>>> q; //dist,a,money
    q.push({0,1,min(s,M)});
    res[1][min(s,M)]=0;
    while(!q.empty())
    {
        auto [dist,a,money]=q.top();
        q.pop();
        if(dist>res[a][money]) continue;
        //cout << "in " << a << " with dist=" << dist << " and money=" << money << endl;
        if(res[a][money]+d[a]<res[a][min(money+c[a],M)])
        {
            res[a][min(money+c[a],M)]=res[a][money]+d[a];
            q.push({res[a][min(money+c[a],M)],a,min(money+c[a],M)});
        }
        for(auto [to,w,t]:v[a])
        {
            if(money<w) continue;
            ll now=money-w;
            if(res[a][money]+t<res[to][now])
            {
                res[to][now]=res[a][money]+t;
                q.push({res[to][now],to,now});
            }
        }
    }
    for(int i=2;i<=n;i++)
    {
        ll best=(1ll<<62);
        for(int j=0;j<=M;j++) best=min(best,res[i][j]);
        printf("%lld\n",best);
    }
    return 0;
}
