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 ll long long
#define mp make_pair
#define pb push_back

#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, n) for (int i = 0; i < (int)(n); ++ i)

struct E {
	int v, a, b;
	E(int _v, int _a, int _b): v(_v), a(_a), b(_b) {}
};

const int mx = 5000;

int n, m, s;
vector <E> g[55];
int C[55], D[55];
int id[55][5555], vu[255555], vs[255555], cnt;
ll dis[255555];

int main() {
	scanf("%d %d %d", &n, &m, &s);
	rep(i, m) {
		int u, v, a, b;
		scanf("%d %d %d %d", &u, &v, &a, &b);
		-- u; -- v;
		g[u].pb(E(v, a, b));
		g[v].pb(E(u, a, b));
	}
	rep(i, n) scanf("%d %d", &C[i], &D[i]);
	rep(i, n) rep(j, mx + 1) {
		id[i][j] = cnt;
		vu[cnt] = i; vs[cnt] = j;
		cnt ++;
	}
	memset(dis, 0x3f, sizeof(dis));
	dis[id[0][min(s, mx)]] = 0;
	priority_queue <pair <ll, int> > pq;
	pq.push(mp(0, id[0][min(s, mx)]));
	while (!pq.empty()) {
		int u = vu[pq.top().second];
		int s = vs[pq.top().second];
		ll d = -pq.top().first;
		pq.pop();
		if (d != dis[id[u][s]]) continue;
		rep(i, g[u].size()) {
			int v = g[u][i].v, a = g[u][i].a, b = g[u][i].b;
			if (s < a) continue;
			int to = id[v][s - a];
			if (dis[to] > d + b) {
				dis[to] = d + b;
				pq.push(mp(-dis[to], to));
			}
		}
		int to = id[u][min(s + C[u], mx)];
		if (dis[to] > d + D[u]) {
			dis[to] = d + D[u];
			pq.push(mp(-dis[to], to));
		}
	}
	rep(u, n) if (u) {
		ll ans = 1e18;
		rep(j, mx + 1) ans = min(ans, dis[id[u][j]]);
		printf("%lld\n", ans);
	}
	return 0;
}
