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 di, Farmer John sends a delivery of bi haybales (1≤di≤1014, 0≤bi≤109).
Process U (1≤U≤105) 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 109+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 109+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≤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