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 pair<int, int> PII;
typedef long long LL;

const LL inf = 0x3f3f3f3f3f3f3f3f;
const int N = 55;
int n, m, s;
LL dis[N][2510];
struct Edge {
	int from, to, next;
	LL w, cost;
}e[N * 2];
int cnt, head[N];
LL c[N], d[N];

void add(int u, int v, int a, int b) {
	e[++cnt].to = v;
	e[cnt].from = u;
	e[cnt].w = b;
	e[cnt].cost = a;
	e[cnt].next = head[u];
	head[u] = cnt;
}

void spfa() {
	queue<PII> q;
	for(int i = 1; i <= n; i++) {
		for(int j = 0; j <= 2500; j++) {
			dis[i][j] = inf;
		}
	}
	dis[1][s] = 0;
	q.push({1, s});
	while(!q.empty()) {
		int u = q.front().first, w = q.front().second; q.pop();
		for(int i = head[u]; i; i = e[i].next) {
			LL v = e[i].to, a = e[i].cost, b = e[i].w;
			if(w >= a && dis[v][w - a] > dis[u][w] + b) {
				dis[v][w - a] = dis[u][w] + b;
				q.push({v, w - a});
			}
		}
		if(dis[u][min(w + c[u], (LL)2500)] > dis[u][w] + d[u]) {
			dis[u][min(w + c[u], (LL)2500)] = dis[u][w] + d[u];
			q.push({u, min(w + c[u], (LL)2500)});
		}
	}
}

void solve() {
	scanf("%d %d %d", &n, &m, &s);
	s = min(s, 2500);
	while(m--) {
		int u, v, a, b;
		scanf("%d%d%d%d", &u, &v, &a, &b);
		add(u, v, a, b);
		add(v, u, a, b);
	}
	for(int i = 1; i <= n; i++) scanf("%lld %lld", &c[i], &d[i]);
	spfa();
//	for(int i = 0; i <= 10; i++) printf("%d\n", dis[2][i]);
	for(int i = 2; i <= n; i++) {
		LL res = inf;
		for(int j = 0; j <= 2500; j++) res = min(res, dis[i][j]);
		printf("%lld\n", res);
	}
}

int main() {
//	freopen("in.txt", "r", stdin);
	solve();
	return 0;
}