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;
typedef long long ll;
typedef long double ld;
#define FOR(i, a, b) for(ll i = (ll)a; i <= (ll)b; i++)
#define DEC(i, a, b) for(ll i = (ll)a; i >= (ll)b; i--)
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
#define f first
#define s second
#define pb push_back
#define all(v) v.begin(), v.end()
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

const ll INF = 1e15;
ll n, m, s, u, v, a, b, dist[55][2505];
vector<pii> adj[55];
priority_queue<pii, vector<pii>, greater<pii> > pq;

int main() {
	fastio; cin >> n >> m >> s;
	s = min(s, 50 * (n-1));
	FOR(i, 1, m) {
		cin >> u >> v >> a >> b;
		adj[u].pb(pii(b, pi(-a, v)));
		adj[v].pb(pii(b, pi(-a, u)));
	}
	FOR(i, 1, n) {
		cin >> a >> b;
		adj[i].pb(pii(b, pi(a, i)));
	}
	memset(dist, -1, sizeof dist);
	pq.push(pii(0, pi(s, 1)));
	dist[1][s] = 0;
	while (!pq.empty()) {
		pii cur = pq.top(); pq.pop();
		for (auto it:adj[cur.s.s]) {
			ll to = it.s.s, nc = cur.s.f + it.s.f, nt = cur.f + it.f;
			nc = min(nc, 50 * (n-1));
			if (nc < 0) continue;
			if (dist[to][nc] == -1 or dist[to][nc] > nt) {
				dist[to][nc] = nt;
				pq.push(pii(nt, pi(nc, to)));
			}
		}
	}
	FOR(i, 2, n) {
		ll ans = INF;
		FOR(j, 0, 50*(n-1)) if (dist[i][j] != -1) ans = min(ans, dist[i][j]);
		cout << ans << "\n";
	}
}
