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Time Limit: 6 s Memory Limit: 512 MB Total points: 100

#5657. Hungry Cow

统计

Note: The time limit for this problem is 6s, three times the default. The memory limit for this problem is 512MB, twice the default.

Bessie is a hungry cow. Each day, for dinner, if there is a haybale in the barn, she will eat one haybale. Farmer John does not want Bessie to starve, so some days he sends a delivery of haybales, which arrive in the morning (before dinner). In particular, on day $d_i$, Farmer John sends a delivery of $b_i$ haybales ($1\leq d_i \leq 10^{14}$, $0\leq b_i \leq 10^9$).

Process $U$ ($1\le U\le 10^5$) updates as follows: Given a pair $(d, b)$, update the number of haybales arriving on day $d$ to $b$. After each update, output the sum of all days on which Bessie eats haybales modulo $10^9+7$.

INPUT FORMAT (input arrives from the terminal / stdin):

$U$, followed by $U$ lines containing the updates.

OUTPUT FORMAT (print output to the terminal / stdout):

The sum after each update modulo $10^9+7$.

SAMPLE INPUT:

3
4 3
1 5
1 2

SAMPLE OUTPUT:

15
36
18
Answers after each update:
4+5+6=15
1+2+3+4+5+6+7+8=36
1+2+4+5+6=18

SAMPLE INPUT:

9
1 89
30 7
101 26
1 24
5 1
60 4
5 10
101 0
1 200

SAMPLE OUTPUT:

4005
4656
7607
3482
3507
3753
4058
1107
24531

SCORING:

  • Input 3: $U\le 5000$
  • Inputs 4-10: Updates only increase the number of haybales arriving on day $d$.
  • Inputs 11-22: No additional constraints.

Problem credits: Brandon Wang and Benjamin Qi

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