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 nl '\n'
typedef long long ll;
const int MAXN = 52, MAXC = 5002;
struct road{
	ll t;
	int v, c;
	friend bool operator<(const road &a, const road &b){
		return a.t>b.t;
	}
};
int N, M, S, sum;
ll dis[MAXN][MAXC];
vector<road>adj[MAXN];
pair<int, ll>trade[MAXN];
priority_queue<road>pq;
int mc(int coins){
	return min(coins, sum);
}
void ist(int curV, int v, int curC, int c, ll curT, ll t){
	if(dis[v][c]>dis[curV][curC]+curT+t){
		dis[v][c] = dis[curV][curC]+curT+t;
		pq.push({dis[v][c], v, c});
	}
}
int main(){
	cin.tie(0);
	cin.sync_with_stdio(0);
	cin >> N >> M >> S;
	for(int i = 1, u, v, a, b;i<=M;i++){
		cin >> u >> v >> a >> b;
		adj[u].push_back({b, v, a});
		adj[v].push_back({b, u, a});
		sum+=a;
	}
	S = mc(S);
	for(int i = 1, c, d;i<=N;i++){
		cin >> c >> d; c = mc(c);
		trade[i] = {c, d};
	}
	memset(dis, 0x3f3f, sizeof(dis));
	dis[1][S] = 0;
	pq.push({0, 1, S});
	while(!pq.empty()){
		road cur = pq.top(); pq.pop();
		int v = cur.v, c = cur.c, cleft = 0;
		for(road i:adj[v]){
			for(int j = 0;j<MAXC;j++){
				cleft = c+(trade[v].first*j)-i.c;
				if(cleft>=sum)break;
				cleft = mc(cleft);
				if(cleft>=0){
					ist(v, i.v, c, cleft, (1LL)*j*trade[v].second, i.t);
				}
			}
		}
	}
	for(int i = 2;i<=N;i++){
		ll qry = 1e18;
		for(int j = 0;j<MAXC;j++){
			qry = min(qry, dis[i][j]);
		}
		cout << qry << nl;
	}
	return 0;
}
