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
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
int n, m, S;
ll dp[100][10000], C[N], D[N];
struct E
{
	int v, a, b;
};
vector<E>G[N];
struct X
{
	ll d;
	int u, c;
	bool operator<(const X& b)const {
		return d > b.d;
	}
};
void bfs(int s) {
	priority_queue<X>q;
	for (int i = 1; i <= n; i++)for (int j = 0; j < 10000; j++)dp[i][j] = inf;
	q.push(X{ 0,s,min(S,6000) });
	dp[s][min(S, 6000)] = 0ll;
	while (!q.empty()) {
		X t = q.top(); q.pop();
		int u = t.u, c = t.c;
		ll d = t.d;
		for (E e : G[u]) {
			if (e.a > c)continue;
			if (dp[e.v][c - e.a] > dp[u][c] + e.b) {
				dp[e.v][c - e.a] = dp[u][c] + e.b;
				q.push(X{ dp[e.v][c - e.a],e.v,c - e.a });
			}
		}
		int nc = min(6000ll, c + C[u]);
		if (dp[u][nc] > dp[u][c] + D[u]) {
			dp[u][nc] = dp[u][c] + D[u];
			q.push(X{ dp[u][nc],u,nc });
		}
	}
}
int main() {
	scanf("%d%d%d", &n, &m, &S);
	for (int i = 0; i < m; i++) {
		int u, v, a, b;
		scanf("%d%d%d%d", &u, &v, &a, &b);
		G[u].push_back(E{ v,a,b });
		G[v].push_back(E{ u,a,b });
	}
	for (int i = 1; i <= n; i++)scanf("%lld%lld", &C[i], &D[i]);
	bfs(1);
	for (int i = 2; i <= n; i++) {
		ll ans = inf;
		for (int j = 0; j <= 6000; j++)
			ans = min(ans, dp[i][j]);
		printf("%lld\n", ans);
	}
	return 0;
}