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;
  
#define int long long 
#define pb push_back
#define S second
#define F first
#define f(i,n) for(int i=0;i<n;i++)
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pii pair<int,int>

const int inf = 1e16;

const int N = 51;
vector<pair<int,pii> > g[N];
int n,m,s;
int c[N];
int d[N];
int dp[N][(N*N)];

void solve()
{
   cin >> n >> m >> s;
    
   int u,v,w,x;
    
   f(i,m)
   {
       cin >> u >> v >> w >> x;
       u--; v--;
       
       g[u].pb({v,{w,x}});
       g[v].pb({u,{w,x}});
   }
    
   f(i,n) cin >> c[i] >> d[i];
   
   if(s >= 2500) s = 2500;
    
   f(i,n) f(j,2501) dp[i][j] = inf; 
   dp[0][s] = 0;
    
  f(go,n+1)  
   for(int i=0;i<n;i++)
   {
       for(auto x : g[i])
           for(int j=0;j<=2500;j++)
       {
         if(j - x.S.F >= 0)  
           dp[i][j - x.S.F] = min(dp[i][j - x.S.F], dp[x.F][j] + x.S.S);
       }
      
       for(int j=0;j<=2500;j++)
       {
           if(j - c[i] >= 0)
           {
                dp[i][j] = min(dp[i][j],dp[i][j-c[i]] + d[i]);
           }
       }
       
   }
    
   for(int i=1;i<n;i++)
   {
       int res = inf;
       
       for(int j=0;j<=2500;j++) res = min(res,dp[i][j]);
       
       cout << res << '\n';
   }
}

signed main()
{
    fast;
    solve();
}