// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define REP(i,a,b) for(int i=(a),_ed=(b);i<=_ed;++i)

#define DREP(i,a,b) for(int i=(a),_ed=(b);i>=_ed;--i)

#define pb(x) push_back((x))

typedef long long ll;

inline int read(){

    register int x=0,f=1;register char ch=getchar();

    while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}

    while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}

    return f?x:-x;

}



const int N=1e5+5;

int n,m,dep[N],s,t,flg,S[N],k[N],stk[N],tp;

ll ans;

vector<int> E[N];



inline void Dfs(int u,int pa){

    for(int v:E[u]){

	if(v==pa)continue;

	if(dep[v]){flg=(dep[u]==dep[v]);s=u,t=v;continue;}

	dep[v]=dep[u]*-1;Dfs(v,u);

    }

}

inline void Cal(int u,int pa){

    for(int v:E[u]){

	if(v==pa||(u==s&&v==t)||(v==s&&u==t))continue;

	Cal(v,u);

	S[u]+=S[v];k[u]+=k[v];

    }

}



int main(){

    //freopen("in.in","r",stdin);

    n=read(),m=read();

    REP(i,1,m){int u=read(),v=read();E[u].pb(v),E[v].pb(u);}

    dep[1]=1;Dfs(1,0);

    int sum=0;

    REP(i,1,n)sum+=(S[i]=dep[i]);

    if(m==n-1){if(sum)return puts("-1"),0;}

    else if((flg&&(sum&1))||(!flg&&sum))return puts("-1"),0;

    if(m==n){

	if(flg){

	    S[s]-=sum/2,S[t]-=sum/2;

	    ans+=abs(sum/2);

	}

	else k[s]=1,k[t]=-1;

    }

    Cal(1,0);

    REP(i,1,n)

	if(!k[i])ans+=abs(S[i]);

	else if(~k[i])stk[++tp]=-S[i];

	else stk[++tp]=S[i];

    stk[++tp]=0;

    sort(stk+1,stk+tp+1);

    int x=stk[(tp+1)>>1];

    REP(i,1,tp)ans+=abs(stk[i]-x);

    printf("%lld\n",ans);

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

// #define int long long

void read(int &x)

{

    int r = 0, ne = 1;

    char c;

    while (c < '0' || c > '9')

    {

        if (c == '-')

            ne = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9')

    {

        r = r * 10 + c - '0';

        c = getchar();

    }

    x = r * ne;

}

struct  Edge

{

    /* data */

    int i,nxt;

};

/* 链表 */

int h[100002],tot;

Edge e[200002];

inline void _add(int x,int y){

    e[++tot].i=y;

    e[tot].nxt=h[x];

    h[x]=tot;

    e[++tot].i=x;

    e[tot].nxt=h[y];

    h[y]=tot;

}

/* 输入 */

int n,m,x,y;

/* 偶环-求解 */

int s[100002],top,mid;

/* dfs */

bool odd;/* 奇环 */

int A,B;/* 环两个连接点 */

int w[100002];/* 点绿色权值(1,-1) */ /* 第二遍扫时存的是子树总权值 */

int k[100002];/* 偶环时标记多余边到LCA的点 */

int sum;/* 权值总和 */

long long ans;/* 答案 */

void dfs(int x,int fa){/* 染色 */

    for(int i=h[x];i;i=e[i].nxt){      

        int o=e[i].i;

        if(o==fa)continue;

        if(w[o]){

            if(w[o]==w[x])

                odd=true;

            A=x,B=o;

        }else{

            w[o]=-w[x];

            // sum+=w[o];

            dfs(o,x);

        }

    }

}

void Dfs(int x,int fa){/* 求解 */

    for(int i=h[x];i!=0;i=e[i].nxt){

        int o=e[i].i;

        if(o==fa||(o==A&&x==B)||(o==B&&x==A))continue;/* 排除基环树的干扰 */

        Dfs(o,x);

        w[x]+=w[o];

        k[x]+=k[o];

    }

}

signed main(){

    read(n),read(m);

    for(register int i=0;i<m;++i){

        read(x),read(y);

        _add(x,y);

    }

    // if(n%2){

    //     puts("-1");

    //     return 0;

    // }

    w[1]=1;

    // sum=1;

    dfs(1,0);

    for(int i=1;i<=n;++i)sum+=w[i];    

    if(m+1==n)/* 树 */{

        if(sum){

            puts("-1");

           return 0;

        }

    }else{

        if(odd)/* 奇环 */{

            if(sum&1){

                puts("-1");

                return 0;

            }

            ans+=abs(sum>>1);

            w[A]-=sum>>1,w[B]-=sum>>1;

        }else/* 偶环 */{

            if(sum){

                puts("-1");

                return 0;

            }else 

                k[A]=1,k[B]=-1;

        }

    }

    Dfs(1,0);

    for(register int i=1;i<=n;++i){

        if(k[i])

            s[++top]=k[i]*w[i];

        else

            ans+=abs(w[i]);

    }

    s[++top]=0;

    sort(s+1,s+top+1);

    mid=s[(top+1)>>1];

    for(register int i=1;i<=top;++i)

        ans+=abs(s[i]-mid);

    printf("%lld\n",ans);

    return 0;

}

// Fastest known solution for this problem: