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
def main():
    from heapq import heappush, heappop

    INF = 1 << 60
    MX_AG = 2500

    N, M, S = map(int, input().split())
    S = min(S, MX_AG)  # それ以上のAgは不要

    g = tuple(set() for _ in range(N))
    for _ in range(M):
        u, v, a, b = map(int, input().split())
        u -= 1
        v -= 1
        g[u].add((v, a, b))
        g[v].add((u, a, b))

    exchange = []
    for _ in range(N):
        c, d = map(int, input().split())
        exchange.append((c, d))

    h = [(0, S, 0)]

    time = [[INF] * (MX_AG + 1) for _ in range(N)]
    time[0][S] = 0
    # time[location][r_Ag]:=minimum_time_to_reach

    while h:
        t, ag, loc = heappop(h)
        for to, fare, dt in g[loc]:
            n_Ag = ag - fare
            if n_Ag < 0: continue
            cf = time[to][n_Ag]
            n_time = t + dt
            if cf <= n_time: continue
            time[to][n_Ag] = n_time
            heappush(h, (n_time, n_Ag, to))

        e_time = t + exchange[loc][1]
        e_Ag = ag + exchange[loc][0]
        if e_Ag > MX_AG: continue
        cf = time[loc][e_Ag]
        if cf <= e_time: continue
        time[loc][e_Ag] = e_time
        heappush(h, (e_time, e_Ag, loc))

    ans = (min(time[to]) for to in range(1, N))

    print('\n'.join(map(str, ans)))


if __name__ == '__main__':
    main()
