### Problem Statement :

```You are given a two-dimensional list of integers reviews and a positive integer threshold. Each element reviews[i] contains [x, y] meaning product i had x number of 5-star reviews and a total of y reviews. All reviews are for one store.

Return the minimum number of additional 5-star reviews we need such that the percentage of 5-star reviews in the store is at least threshold. You can assume that it's possible to reach threshold% of 5-star reviews.

Constraints

1 ≤ n ≤ 100,000 where n is the length of reviews

0 ≤ threshold ≤ 100

Example 1

Input

reviews = [
[4, 4],
[1, 2],
[3, 6]
]

threshold = 77

Output

6

Explanation

So in total there were 8 5-star reviews and a total of 12 reviews. To reach 77% 5-star reviews, we need 6
more 5-star reviews.

Example 2

Input
reviews = [
[10, 20]
]

threshold = 50

Output

0

Explanation

We're already at 50% 5-star reviews.

Example 3

Input

reviews = [
[1, 1]
]

threshold = 100

Output

0

Explanation

We're already at 100% 5-star reviews.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& reviews, int threshold) {
double fi = 0, xi = 0;
for (int i = 0; i < reviews.size(); i++) {
fi += reviews[i];
xi += reviews[i];
}

return max((double)0,
ceil((threshold * fi - 100 * xi) / max((double)1, (100 - (double)threshold))));
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[][] reviews, int threshold) {
if (threshold == 0)
return 0;
int totalReview = 0;
int total5Star = 0;
for (int i = 0; i < reviews.length; i++) {
totalReview += reviews[i];
total5Star += reviews[i];
}

// (totalReview + x) * threshold / 100 = (total5Star + x)

int res = (int) Math.ceil(
(double) ((totalReview * threshold) - (100 * total5Star)) / (100 - threshold));

return res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, reviews, threshold):
a = 0
b = 0
for c, d in reviews:
a += c
b += d
if a * 100 >= threshold * b:
return 0
delta = threshold * b - 100 * a
return (delta + (99 - threshold)) // (100 - threshold)```
```

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