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 int long long
inline void read (int &x) {
	char ch = getchar(); x = 0;
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
} const int N = 55, M = 210;
int n, m, s, c[N], d[N], f[N][N * N];
int cnt, h[N], to[M], nxt[M], a[M], b[M];
inline void add (int u, int v, int x, int y) {
	to[++cnt] = v, a[cnt] = x, b[cnt] = y, nxt[cnt] = h[u], h[u] = cnt;
}
struct node {
	int res, u, s;
	inline bool operator < (const node &x) const { return res > x.res; }
} ; priority_queue<node> q;
inline void Dij() {
	q.push (node {0, 1, s});
	while (!q.empty()) {
		node t = q.top(); q.pop();
		if (f[t.u][t.s] < t.res) continue;
		for (int i = h[t.u]; i; i = nxt[i])
			if (t.s >= a[i] && f[to[i]][t.s - a[i]] > t.res + b[i])
				f[to[i]][t.s - a[i]] = t.res + b[i], q.push (node {t.res + b[i], to[i], t.s - a[i]});
		int qwq = min (2500ll, t.s + c[t.u]);
		if (f[t.u][qwq] > t.res + d[t.u])
			f[t.u][qwq] = t.res + d[t.u], q.push (node {t.res + d[t.u], t.u, qwq});
	}
}
signed main() {
	read (n), read (m), read (s); s = min (s, 2500ll);
	memset (f, 0x3f, sizeof (f)); f[1][s] = 0;
	for (int i = 1, u, v, x, y; i <= m; ++i) {
		read (u), read (v), read (x), read (y);
		add (u, v, x, y), add (v, u, x, y);
	} for (int i = 1; i <= n; ++i) read (c[i]), read (d[i]); Dij (); 
	for (int i = 2; i <= n; ++i) {
		int res = f[0][0];
		for (int j = 0; j <= 2500; ++j) res = min (res, f[i][j]);
		printf ("%lld\n", res);
	} return 0;
}