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 solve():
    import sys
    input = sys.stdin.readline
    N,M,S = map(int,input().split())
    UVAB = [tuple(map(int,input().split())) for i in range(M)]
    CD = [tuple(map(int,input().split())) for i in range(N)]

    es = [[] for _ in range(N)]
    for u,v,a,b in UVAB:
        u,v = u-1,v-1
        es[u].append((v,a,b))
        es[v].append((u,a,b))

    INF = float('inf')
    times = [INF] * (2501 * N)
    start = min(2500,S)*N
    times[start] = 0
    hq = [start]
    ans = [-1] * N
    ans[0] = 0
    reached = 0
    import heapq
    heapq.heapify(hq)
    while hq:
        x = heapq.heappop(hq)
        t,v = divmod(x,N*2501)
        s,i = divmod(v,N)
        if ans[i] < 0:
            ans[i] = t
            reached += 1
            if reached == N-1: break
        for to,a,b in es[i]:
            if a > s: continue
            nv = (s-a)*N + to
            if times[nv] <= t + b: continue
            times[nv] = t + b
            heapq.heappush(hq, (t+b)*N*2501 + nv)
        ns = s
        nt = times[v]
        while ns <= 2500:
            ns += CD[i][0]
            nt += CD[i][1]
            nv = min(2500,ns)*N + i
            if times[nv] <= nt: break
            times[nv] = nt
            heapq.heappush(hq, nt*N*2501 + nv)
    print(*ans[1:], sep='\n')
solve()