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 pb emplace_back
using namespace std;
using ll = long long;
const int maxn = 51, maxm = maxn * maxn;
const ll inf = 1ll << 55;
int n, m, s, gain[maxn], wait[maxn];
ll dp[maxn][maxm];
bool vis[maxn][maxm];
void init() {
	for (int i = 0;i <= n;++i)
		fill(dp[i], dp[i]+maxm, inf);
}
struct info { int u, a, w; };
vector<info> edge[maxn];
#define NW(i) min(i, maxm-1)
struct st {
	int now, mon; ll len;
	bool operator < (const st &b) const {
		return len > b.len;
	}
};
void dij() {
	// [len, now]
	priority_queue<st> pq;
	pq.push({1, s, 0});
	while (pq.size()) {
		auto [now, mon, len] = pq.top();
		pq.pop();
		if (vis[now][mon]) continue;
		vis[now][mon] = true;
		for (auto [u, a, w] : edge[now]) if (mon >= a && dp[u][mon-a] > len + w) {
			dp[u][mon-a] = len + w;
			pq.push({u, mon-a, len+w});
		}
		if (len + wait[now] < dp[now][NW(mon + gain[now])]) {
			dp[now][NW(mon + gain[now])] = len + wait[now];
			pq.push({now, NW(mon+gain[now]), len+wait[now]});
		}
	}
	for (int i = 2;i <= n;++i)
		cout << *min_element(dp[i], dp[i]+maxm) << '\n';
}
signed main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m >> s;
	for (int i = 0, u, v, a, b;i < m;++i) {
		cin >> u >> v >> a >> b;
		edge[u].push_back({v, a, b});
		edge[v].push_back({u, a, b});
	}
	for (int i = 1;i <= n;++i)
		cin >> gain[i] >> wait[i];
	init();
	s = min(s, maxm-1);
	dij();
}
