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 pair<ll, ll> P;
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define vcout(v) rep(i,v.size()) cout << v[i] << " "; cout << endl
#define mcout(m) rep(j,m.size()){vcout(m[j]);}
const ll MOD = 1e9 + 7;

struct Info{ll v; ll c; ll t;};
struct DP{ll v; ll c; ll t;};

int mainsolve(){

	ll n,m,s;
	cin >> n >> m >> s;
	vector<ll> a(m), b(m);
	vector<vector<Info>> to(n);
	rep(i,m){
		ll u,v;
		cin >> u >> v >> a[i] >> b[i];
		--u; --v;
		to[u].push_back(Info{v,a[i],b[i]});
		to[v].push_back(Info{u,a[i],b[i]});
	}
	vector<ll> c(n), d(n);
	rep(i,n) cin >> c[i] >> d[i];

	ll INF = 1e18;
	ll maxc = n*50; // 銀貨の枚数はこれだけあれば十分
	vector<vector<ll>> dp(n,vector<ll>(maxc,INF)); // dp[v][c]: 銀貨c枚持ってvにいる状態にするためにかかる時間 

	queue<DP> q;
	q.push(DP{0,min(maxc-1,s),0});

	while(!q.empty()){
		DP now = q.front(); q.pop();
		if(now.c >= 0 && now.t >= 0 && now.t < dp[now.v][min(maxc-1,now.c)] ){
			dp[now.v][min(maxc-1,now.c)] = now.t;

			for(Info next : to[now.v]){
				q.push(DP{next.v,now.c - next.c, now.t + next.t}); // 移動
			}
			q.push(DP{now.v,now.c + c[now.v], now.t + d[now.v]}); // 換金
		}
	}

	rep(i,n){
		if(i==0) continue;
		ll res = INF;
		rep(j,maxc) res = min(res, dp[i][j]);
		cout << res << endl;
	}
	
	return 0;
}

int main(){

	return mainsolve();
}