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++)
using ll = long long;
using T = tuple<int, int, int>;
using T2 = tuple<ll, int, int>;
constexpr ll INF = 1e18+7;
int n, m, s, u, v, a, b;
vector<T> tp[55];
ll dp[55][2505], c[55], d[55], ans = INF;
priority_queue<T2, vector<T2>, greater<T2>> qu;
int main(){
	scanf("%d%d%d", &n, &m, &s);
	s = min(2499, s);
	fill(dp[0], dp[50], INF);
	rep(i,m){
		scanf("%d%d%d%d", &u, &v, &a, &b);
		u--; v--;
		tp[u].emplace_back(v, a, b);
		tp[v].emplace_back(u, a, b);
	}
	rep(i,n) scanf("%lld%lld", c+i, d+i);
	qu.push({dp[0][s] = 0, 0, s});
	while(!qu.empty()){
		auto[cost, town, coin] = qu.top(); qu.pop();
		if(dp[town][coin] < cost) continue;
		for(auto [nxtown, coin2, cost2]: tp[town]){
			if(coin2 > coin) continue;
			if(dp[nxtown][coin - coin2] > cost + cost2){
				qu.push({dp[nxtown][coin - coin2] = cost + cost2, nxtown, coin - coin2});
			}
		}
		if(coin + c[town] < 2500){
			if(dp[town][coin + c[town]] > cost + d[town]){
				qu.push({dp[town][coin + c[town]] = cost + d[town], town, coin + c[town]});
			}
		}
	}
	for(int i=1; i<n; i++){
		ans = INF;
		rep(j,2500) ans = min(ans, dp[i][j]);
		printf("%lld\n", ans);
	}
}