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 <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#define N 55
using namespace std;
typedef long long ll;

struct st{ll ct, co, tm;};
ll n, m, s, c[N], d[N], v[N][5005], ans[N];
vector<st> gr[N];
priority_queue<st> pq;

bool operator < (st p, st q) {return p.tm > q.tm;}

int main() {
	ll i, t1, t2, t3, t4;
	cin >> n >> m >> s;
	while(m--) {
		cin >> t1 >> t2 >> t3 >> t4;
		gr[t1].push_back({t2, t3, t4});
		gr[t2].push_back({t1, t3, t4});
	}
	for(i=1; i<=n; i++) cin >> c[i] >> d[i];
	pq.push({1, min(s, 5000LL), 0});
	while(pq.size()) {
		st t = pq.top(); pq.pop();
		if(v[t.ct][t.co]) continue;
		v[t.ct][t.co] = t.tm;
		if(!ans[t.ct]) ans[t.ct] = t.tm;
		for(auto x:gr[t.ct]) {
			if(t.co>=x.co && !v[x.ct][t.co-x.co]) {
				pq.push({x.ct, t.co-x.co, t.tm+x.tm});
			}
		}
		if(t.co+c[t.ct] <= 5000) pq.push({t.ct, t.co+c[t.ct], t.tm+d[t.ct]});
	}
	for(i=2; i<=n; i++) cout << ans[i] << endl;
 	return 0;
}