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 rep(i,n) for(int i=0;i<(n);++i)
#define Rep(i,n) for(int i=1;i<=(n);++i)
using LL = long long;
using ULL = unsigned long long;

const LL INF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = 10000;

vector<vector<int>> e[53];
int c[53], d[53];
LL ans[53]{0, 1}, dis[51][10001];

int main()
{
	int n, m, s;
	cin >> n >> m >> s;
	rep(i, m) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		e[u].push_back({v, a, b});
		e[v].push_back({u, a, b});
	}
	Rep(i, n) cin >> c[i] >> d[i];
	priority_queue<vector<LL>> q;
	memset(dis, 0x3f, sizeof dis);
	s = min(s, MAX);
	q.push({0, -s, 1});
	dis[1][s] = 0;
	int cnt = n - 1;
	while(true) {
		auto x = q.top(); q.pop();
		int p = x[2];
		if(ans[p] == 0) {
			ans[p] = -x[0];
			if(--cnt == 0) break;
		}
		if(dis[p][-x[1]] < -x[0]) continue;
		LL coin = min(c[p]-x[1], (LL)MAX), t = d[p] - x[0];
		if(t < dis[p][coin]) {
			q.push({-t, -coin, p});
			dis[p][coin] = t;
		}
		for(auto& e0: e[p]) {
			coin = -(x[1] + e0[1]); t = e0[2] - x[0];
			if(coin >= 0 && t < dis[e0[0]][coin]) {
				q.push({-t, -coin, e0[0]});
				dis[e0[0]][coin] = t;
			}
		}
	}
	for(int i = 2; i <= n; ++i) cout << ans[i] << endl;
	return 0;
}
