For problem statement at 1000-1999/1700-1799/1700-1709/1700/problemD.txt this is a correct solution, but verifier at 1000-1999/1700-1799/1700-1709/1700/verifierD.go ends with All 100 tests passed. can you fix the verifier? Looking at this problem, I need to find the minimum number of pipes to open so that all locks are filled within time `t`.

Key insight: If I open pipes at positions `p1 < p2 < ... < pk`, each pipe contributes `t` liters of water over `t` seconds. Water flows forward (from lock i to i+1 when i overflows). So opening a pipe at position `i` adds `t` liters starting at lock `i`, filling locks from `i` onward.

To minimize pipes, I should use a greedy approach: open pipes as late as possible (rightmost first won't work easily), or think of it as: we need to cover all locks with water. Opening a pipe at position `i` provides `t` liters that fill locks `i, i+1, i+2, ...` in order.

**Greedy from left to right**: We must fill lock 1. The only way is to open a pipe at lock 1 (or earlier - lock 1 is the first). Actually, a pipe at lock `j ≤ i` can contribute to lock `i` if there's enough overflow. 

So greedily: scan left to right. We need the total water from all opened pipes to cover the cumulative volume. Open a pipe as late as possible.

Actually, let me think differently. Let `S[i] = v[1] + v[2] + ... + v[i]` (prefix sums). If we open `k` pipes at positions `p1 ≤ p2 ≤ ... ≤ pk`, then the total water available for locks 1..i is `(number of pipes at positions ≤ i) * t`. For all locks to be filled, we need for every `i`: `(count of pipes at positions ≤ i) * t ≥ S[i]`.

So the minimum number of pipes at positions ≤ i must be `ceil(S[i] / t)`. We need `ceil(S[i]/t) ≤ i` for all `i`, otherwise it's impossible. The total number of pipes is `ceil(S[n]/t)`, provided `ceil(S[i]/t) ≤ i` for all `i`.

Wait - is it that simple? If `ceil(S[i]/t) ≤ i` for all `i`, then we can greedily assign pipes. The answer is `ceil(S[n]/t)` if feasible, else -1.

Feasibility: for each `i`, `ceil(S[i]/t) ≤ i`, i.e., `S[i] ≤ i * t`. The bottleneck is `max over i of ceil(S[i]/t)` must equal at most `i` for each respective `i`. If any `ceil(S[i]/t) > i`, it's -1.

Actually the answer is `ceil(S[n]/t)` when feasible. Feasibility requires `ceil(S[i]/t) ≤ i` for all `i` from 1 to n.

For a given `t`, check: is `S[i] > i * t` for any `i`? If yes, -1. Otherwise answer is `ceil(S[n]/t)`.

The condition `S[i] > i*t` ↔ `S[i]/i > t` ↔ `t < S[i]/i`. So it's impossible iff `t < max(S[i]/i)` (using ceiling division: `t < ceil(S[i]/i)` in integer terms).

More precisely: impossible when there exists `i` with `S[i] > i * t`, i.e., `t < ceil(S[i]/i)`. So impossible iff `t < max over i of ceil(S[i]/i)`.

Let me precompute `T_min = max over i=1..n of ceil(S[i]/i)`. If `t < T_min`, answer is -1. Otherwise answer is `ceil(S[n]/t)`.

```go
package main

import (
	"bufio"
	"fmt"
	"os"
)

func main() {
	reader := bufio.NewReader(os.Stdin)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()

	var n int
	fmt.Fscan(reader, &n)

	v := make([]int64, n)
	for i := 0; i < n; i++ {
		fmt.Fscan(reader, &v[i])
	}

	// Compute prefix sums and T_min
	S := make([]int64, n+1)
	var Tmin int64 = 0
	for i := 1; i <= n; i++ {
		S[i] = S[i-1] + v[i-1]
		// ceil(S[i] / i)
		ii := int64(i)
		c := (S[i] + ii - 1) / ii
		if c > Tmin {
			Tmin = c
		}
	}

	totalVol := S[n]

	var q int
	fmt.Fscan(reader, &q)
	for ; q > 0; q-- {
		var t int64
		fmt.Fscan(reader, &t)
		if t < Tmin {
			fmt.Fprintln(writer, -1)
		} else {
			ans := (totalVol + t - 1) / t
			fmt.Fprintln(writer, ans)
		}
	}
}
```