There is a platform with n columns. 1×1 squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.

When all of the n columns have at least one square in them, the bottom row is being removed. You will receive 1 point for this, and all the squares left will fall down one row.

You task is to calculate the amount of points you will receive.

The first line of input contain 2 integer numbers n and m (1≤n,m≤1000) — the length of the platform and the number of the squares.

The next line contain mm integer numbers c1,c2,…,cm (1≤ci≤n) — column in which i-th square will appear.

Print one integer — the amount of points you will receive.

In the sample case the answer will be equal to 2 because after the appearing of 6-th square will be removed one row (counts of the squares on the platform will look like [2 3 1], and after removing one row will be [1 2 0]).

After the appearing of 9-th square counts will be [2 3 1], and after removing one row it will look like [1 2 0].

【大意及分析】

有点像俄罗斯方块，大意就是在n列的区域中，随机在给定的若干列依次出现m个小方块，每个小方块都会占据本列空闲位置的最下方，当每一列的最后一行都有小方块时，这一行就会被删掉，分数加一，然后上面的方块下落一行。

我们只需要利用数组表示每一列的方块个数即可（注意不要用二维数组表示每一列的方块的情况，我们并不需要知晓方块的具体位置，那样算法复杂度会变大）。

【示例代码】

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #define MAX_N 1005
 4 int a[MAX_N];
 5
 6 int main() {
 7     int score = 0;
 8     int n, m, temp, bottom_score = 0;
 9     bool judge;
10     scanf("%d %d", &n, &m);
11     for (int i = 0; i < n; i++)
12         a[i] = 0;
13     for (int i = 0; i < m; i++) {
14         scanf("%d", &temp);
15         a[temp - 1]++;
16         judge = true;
17         for (int k = 0; k < n; k++) {
18             if (a[k] == bottom_score) {
19                 judge = false;
20                 break;
21             }
22         }
23         if (judge) {
24             score++;
25             bottom_score++;
26         }
27     }
28     printf("%d\n", score);
29     return 0;
30 }