Codeforces 556C Case of Matryoshkas【模拟】

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Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of contemporary art.

The main exhibit is a construction of n matryoshka dolls that can be nested one into another. The matryoshka dolls are numbered from1 to n. A matryoshka with a smaller number can be nested in a matryoshka with a higher number, two matryoshkas can not be directly nested in the same doll, but there may be chain nestings, for example, 1 → 2 → 4 → 5.

In one second, you can perform one of the two following operations:

Having a matryoshka a that isn't nested in any other matryoshka and a matryoshkab, such that b doesn't contain any other matryoshka and is not nested in any other matryoshka, you may puta in b;
Having a matryoshka a directly contained in matryoshkab, such that b is not nested in any other matryoshka, you may geta out of b.

According to the modern aesthetic norms the matryoshka dolls on display were assembled in a specific configuration, i.e. as several separate chains of nested matryoshkas, but the criminal, following the mysterious plan, took out all the dolls and assembled them into a single large chain (1 → 2 → ... → n). In order to continue the investigation Andrewid needs to know in what minimum time it is possible to perform this action.

Input

The first line contains integers n (1 ≤ n ≤ 10⁵) andk (1 ≤ k ≤ 10⁵) — the number of matryoshkas and matryoshka chains in the initial configuration.

The next k lines contain the descriptions of the chains: thei-th line first contains number m_i (1 ≤ m_i ≤ n), and thenm_i numbersa_i1, a_i2, ..., a_{im_i} — the numbers of matryoshkas in the chain (matryoshka a_i1 is nested into matryoshkaa_i2, that is nested into matryoshkaa_i3, and so on till the matryoshkaa_{im_i} that isn't nested into any other matryoshka).

It is guaranteed that m₁ + m₂ + ... + m_k = n, the numbers of matryoshkas in all the chains are distinct, in each chain the numbers of matryoshkas follow in the ascending order.

Output

In the single line print the minimum number of seconds needed to assemble one large chain from the initial configuration.

Examples

Input

3 22 1 21 3

Output

Input

7 33 1 3 72 2 52 4 6

Output

Note

In the first sample test there are two chains: 1 → 2 and3. In one second you can nest the first chain into the second one and get1 → 2 → 3.

In the second sample test you need to disassemble all the three chains into individual matryoshkas in 2 + 1 + 1 = 4 seconds and then assemble one big chain in 6 seconds.

题目大意：

一共有N个娃娃，我们的目的就是将所有的娃娃串在一起。

我们需要保证从左到右的娃娃的编号是递增的才行。

我们有两种操作：

①拆出当前串中最大编号的娃娃（且一定是最右边的娃娃）。

②连接一个单个的娃娃（不能和其他娃娃相连着的娃娃）。

问最少操作次数，使得所有娃娃串在一起。

思路：

1、根据题目要求可知，如果当前这一串娃娃的起点不是 1，那么显然这一整串是一定需要拆开的，操作次数为：这一串娃娃的个数-1.

如果当前这一串娃娃的起点是1，但是之后的娃娃不是以a【i】-a【i-1】==1的形式递增的，那么后边的部分也要拆开。

2、将所有的娃娃都拆开了之后，我们再将每个娃娃一个一个的搭上去即可。

Ac代码：

#include<stdio.h>#include<queue>#include<string.h>using namespace std;int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        int output=0;        int len=0;        for(int i=0;i<m;i++)        {            int k;            int flag=0;            scanf("%d",&k);            for(int j=0;j<k;j++)            {                int x;                scanf("%d",&x);                if(j==0&&x==1)flag=1,len=1;                if(j>0&&flag==0)output++;                if(j>0&&flag==1)                {                    if(j==x-1)                    {                        len++;                    }                    else                    {                        flag=0;                        output++;                    }                }            }        }        printf("%d\n",output+n-len);    }}

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