// #pragma GCC target("avx2")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-10;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 3121000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
}
// ここから編集しろ
int p[110000];
int t[110000];
int main(){
	int a,b;scanf("%d%d",&a,&b);
	for(int i=0;i<a;i++)p[i]=i;
	int at=0;
	t[0]=1;
	t[1]=1;
	for(int i=0;i<b;i++){
		int x,y;scanf("%d%d",&x,&y);x--;y--;
		if(at==x){
			at=y;
		}else if(at==y){
			at=x;
		}
		swap(p[x],p[y]);
		if(at)t[p[at-1]]=1;
		if(at+1<a)t[p[at+1]]=1;
	}
	int ret=0;
	for(int i=0;i<a;i++)ret+=t[i];
	printf("%d\n",ret);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

You have N cups and 1 ball.

The cups are arranged in a row, from left to right.

You turned down all the cups, then inserted the ball into the leftmost cup.

Then, you will perform the following Q operations:

* The i-th operation: swap the positions of the A_i-th and B_i-th cups from the left. If one of these cups contains the ball, the ball will also move.



Since you are a magician, you can cast a magic described below:

* Magic: When the ball is contained in the i-th cup from the left, teleport the ball into the adjacent cup (that is, the (i-1)-th or (i+1)-th cup, if they exist).



The magic can be cast before the first operation, between two operations, or after the last operation, but you are allowed to cast it at most once during the whole process.

Find the number of cups with a possibility of containing the ball after all the operations and possibly casting the magic.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq A_i < B_i \leq N

Input

The input is given from Standard Input in the following format:


N Q
A_1 B_1
A_2 B_2
:
A_Q B_Q


Output

Print the number of cups with a possibility of eventually containing the ball.

Examples

Input

10 3
1 3
2 4
4 5


Output

4


Input

20 3
1 7
8 20
1 19


Output

5

### Response

```cpp
// #pragma GCC target("avx2")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-10;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 3121000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
}
// ここから編集しろ
int p[110000];
int t[110000];
int main(){
	int a,b;scanf("%d%d",&a,&b);
	for(int i=0;i<a;i++)p[i]=i;
	int at=0;
	t[0]=1;
	t[1]=1;
	for(int i=0;i<b;i++){
		int x,y;scanf("%d%d",&x,&y);x--;y--;
		if(at==x){
			at=y;
		}else if(at==y){
			at=x;
		}
		swap(p[x],p[y]);
		if(at)t[p[at-1]]=1;
		if(at+1<a)t[p[at+1]]=1;
	}
	int ret=0;
	for(int i=0;i<a;i++)ret+=t[i];
	printf("%d\n",ret);
}

```