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>
#define ll long long
#define maxn 55
#define limit 2500
using namespace std;
struct edge {
	int v,a,b;
};
vector<edge> G[maxn];
int value[maxn],cost[maxn];
struct state {
	int u,num;
	ll tim;
	bool operator < (const state& a) const{
		return tim > a.tim;
	}
};
ll dis[maxn][maxn*maxn];
void dijkstra(int s) {
	s = min(s,limit);
	
	memset(dis,-1,sizeof dis);
	dis[1][s] = 0;
	priority_queue<state> pq;
	pq.push(state{1,s,0});
	
	while(!pq.empty()) {
		state now = pq.top();
		pq.pop();
		
		int u = now.u , num = now.num;
		ll tim = now.tim;
		if(dis[u][num] != tim)	continue;
		
		for(edge e : G[u]) {
			int v = e.v , a = e.a , b = e.b;
			if(num >= a) {
				if(dis[v][num - a] == -1 || dis[v][num - a] > tim + b) {
					dis[v][num - a] = tim + b;
					pq.push(state{v,num - a,tim + b});
				}
			}
		}
		
		num += value[u] , tim += cost[u];
		if(num <= limit && (dis[u][num] == -1 || dis[u][num] > tim)) {
			dis[u][num] = tim;
			pq.push(state{u,num,tim});
		}
	}
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,m,s;
	cin >> n >> m >> s;
	while(m--) {
		int u,v,a,b;
		cin >> u >> v >> a >> b;
		G[u].push_back(edge{v,a,b});
		G[v].push_back(edge{u,a,b});
	}
	for(int i=1;i<=n;i++)
		cin >> value[i] >> cost[i];
	
	dijkstra(s);
	
	for(int i=2;i<=n;i++) {
		ll ans = -1;
		for(int j=0;j<=limit;j++)
			if(dis[i][j] != -1 && (ans == -1 || ans > dis[i][j]))
				ans = dis[i][j];
		cout << ans << endl;
	}
	return 0;
}