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(i, 0, (n))
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define LAR(a, b) ((a)=max((a),(b)))
#define SML(a, b) ((a)=min((a),(b)))
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;
template<typename T>
using pque = priority_queue<T, vector<T>, greater<T>>;
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define ALL(a) (a).begin(), (a).end()
#ifdef LOCAL_DEBUG
#define DEBUG(...) printf(__VA_ARGS__)
#else
#define DEBUG(...)
#endif

constexpr int N = 64;
constexpr ll M = 2500;
vector<tuple<int, ll, ll>> g[N];
ll ans[N];
ll c[N], d[N];
bool u[N][M+1] = {};

pque<tuple<ll, int, ll>> que;

int main(){
	int n, m; ll s; scanf("%d%d%lld", &n, &m, &s);
	REP(i, m){
		int u, v; ll a, b; scanf("%d%d%lld%lld", &u, &v, &a, &b);
		g[u].EB(v, a, b);
		g[v].EB(u, a, b);
	}
	REP(i, n) scanf("%lld%lld", c+i+1, d+i+1);

	fill(ans, ans+N, -1);
	que.emplace(0, 1, min(s, M));
	while(que.size()){
		auto [a, i, s] = que.top(); que.pop();
		if(u[i][s]) continue;
		u[i][s] = true;
		if(!~ans[i]) ans[i] = a;
		// DEBUG("%d: %lld, %lld\n", i, a, s);
		// DEBUG("   => %d: %lld, %lld\n", i, a+d[i], min(s+c[i], M));
		que.emplace(a+d[i], i, min(s+c[i], M));
		for(auto [j, t, b] : g[i]){
			if(t > s) continue;
			// DEBUG("   => %d: %lld, %lld\n", j, a+b, s-t);
			que.emplace(a + b, j, s - t);
		}
	}

	FOR(i, 2, n+1) printf("%lld\n", ans[i]);
}



