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 vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define FOR(i,a,b) for (ll i = (a); i < (b); i++)
#define FORD(i,a,b) for (ll i = (b)-1; i >= (a); i--)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define TR(X) ({ if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll n, m, s;
	cin >> n >> m >> s;

	ll k = 2525;
	s = min(k,s);

	vl u(m), v(m), a(m), b(m);
	vvl adj(n);
	FOR(i,0,m) {
		cin >> u[i] >> v[i] >> a[i] >> b[i];
		u[i]--, v[i]--;
		adj[u[i]].pb(i);
		adj[v[i]].pb(i);
	}

	
	vl c(n), d(n);
	FOR(i,0,n) cin >> c[i] >> d[i];
	
	vvl dist(n, vl(k+1, oo));
	set<pair<ll,pll>> q;

	dist[0][s] = 0;
	q.insert({0, {0,s}});

	while (sz(q)) {
		ll i, j;
		tie(i,j) = begin(q)->yy;
		q.erase(begin(q));

		auto aug = [&](ll ni, ll nj, ll nd) {
			if (nd >= dist[ni][nj]) return;
			q.erase({dist[ni][nj], {ni,nj}});
			dist[ni][nj] = nd;
			q.insert({dist[ni][nj], {ni,nj}});
		};

		aug(i, min(k, j+c[i]), dist[i][j] + d[i]);
		
		for (ll t: adj[i]) {
			if (a[t] > j) continue;
			ll ni = i ^ u[t] ^ v[t];
			aug(ni, j-a[t], dist[i][j] + b[t]);
		}
	}
	
	FOR(i,1,n) cout << *min_element(all(dist[i])) << endl;
}

