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 rep(i,a,b) for(int i = a; i < b; ++i)
#define all(c) c.begin(), c.end()
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define gadd(x,y) x=add(x,y)
#define gmul(x,y) x=mul(x,y)
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;
typedef vector<int> vi;

const int N = 50;

vector<pair<int,pii>> g[N];
bool vis[N][N*N+1];
ll d[N][N*N+1];
int cnt[N], cst[N];

void dij(int s){
	priority_queue<pair<ll,pii>, vector<pair<ll,pii>>, greater<pair<ll,pii>>> pq;	
	pq.push({0,{0,s}});
	while(pq.size()){
		ll dist = pq.top().first;
		int u = pq.top().second.first;
		int c = pq.top().second.second;
		pq.pop();
		if(!vis[u][c]){
			vis[u][c] = 1;
			d[u][c] = dist;
			if(c != N*N && !vis[u][c+1]){
				pq.push({dist + cst[u],{u,min(c+cnt[u],N*N)}});
			}
			for(auto e: g[u]){
				int to = e.first;	
				int coin_cst = e.second.first;
				int w = e.second.second;
				if(c >= coin_cst && !vis[to][c-coin_cst]){
					pq.push({dist + w,{to,c-coin_cst}});	
				}
			}
		}
	}
}

int main(){
	cin.tie(0);
	cout.tie(0);
	ios_base::sync_with_stdio(0);
	int n,m,s;
	cin >> n >> m >> s;
	rep(i,0,m){
		int x,y,coins,w;
		cin >> x >> y >> coins >> w;
		--x,--y;
		g[x].push_back({y,{coins,w}});
		g[y].push_back({x,{coins,w}});
	}
	rep(i,0,n)cin >> cnt[i] >> cst[i];
	dij(min(s,N*N));
	rep(i,1,n)rep(j,0,N*N+1)if(vis[i][j])gmin(d[i][0],d[i][j]);
	rep(i,1,n)cout << d[i][0] << '\n';
}
