You are given an `m x n`

matrix `M`

initialized with all `0`

's and an array of operations `ops`

, where `ops[i] = [a`

means _{i}, b_{i}]`M[x][y]`

should be incremented by one for all `0 <= x < a`

and _{i}`0 <= y < b`

._{i}

Count and return *the number of maximum integers in the matrix after performing all the operations*.

**Example 1:**

Input:m = 3, n = 3, ops = [[2,2],[3,3]]Output:4Explanation:The maximum integer in M is 2, and there are four of it in M. So return 4.

**Example 2:**

Input:m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3]]Output:4

**Example 3:**

Input:m = 3, n = 3, ops = []Output:9

**Constraints:**

`1 <= m, n <= 4 * 10`

^{4}`1 <= ops.length <= 10`

^{4}`ops[i].length == 2`

`1 <= a`

_{i}<= m`1 <= b`

_{i}<= n

```
struct Solution;
impl Solution {
fn max_count(mut m: i32, mut n: i32, ops: Vec<Vec<i32>>) -> i32 {
for op in ops {
m = i32::min(op[0], m);
n = i32::min(op[1], n);
}
m * n
}
}
#[test]
fn test() {
let m = 3;
let n = 3;
let ops: Vec<Vec<i32>> = vec_vec_i32![[2, 2], [3, 3]];
let res = 4;
assert_eq!(Solution::max_count(m, n, ops), res);
}
```