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 rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, pll> pl3;

const ll max_n = 52;
const ll max_s = 2500;
vector<ll> c, d;
vector<pl3> g[max_n];

ll n, m;
ll s;

const ll INF = 1e15;
ll dp[max_n+1][max_s + 1];

void dijkstra(){
	s = min(s, max_s);
	rep(i, n) rep(j, max_s + 1) dp[i][j] = INF;
	dp[0][s] = 0;

	priority_queue<pl3,vector<pl3>,greater<pl3> > que;
	que.push({0, {0, s}});

	while(!que.empty()){
		pl3 p = que.top();
		que.pop();

		ll time = p.first;
		ll v = p.second.first;
		ll ns = p.second.second;

		if(dp[v][ns] < time) continue; //後から最短パスが見つかった場合、先にqueに入っているものはここではじかれる

		if(ns + c[v] <= max_s && dp[v][ns + c[v]] > time + d[v]){
			dp[v][ns + c[v]] = time + d[v];
			que.push({time + d[v], {v, ns + c[v]}});
		}

		for(pl3 x : g[v]){
			ll w = x.first;
			ll a = x.second.first;
			ll b = x.second.second;

			if(ns < a || dp[w][ns - a] <= time + b) continue;
			dp[w][ns - a] = time + b;
			que.push({time + b, {w, ns - a}});
		}
	}
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n >> m >> s;
	rep(i, m){
		ll u, v, a, b;
		cin >> u >> v >> a >> b;
		u--;
		v--;
		g[u].push_back({v, {a, b}});
		g[v].push_back({u, {a, b}});
	}

	c.resize(n);
	d.resize(n);
	rep(i, n) cin >> c[i] >> d[i];

	dijkstra();

	rep(i, n-1){
		ll ans = INF;
		rep(j, max_s + 1) ans = min(ans, dp[i+1][j]);
		cout << ans << "\n";
	}
}
