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

using namespace std;

const int N = 100010;

vector<int> G[N];

int delv, delu;

int n, m, x, y, d[N], fa[N], dp[N], q[N];

void dfs(int x, int fx) {

    d[x] = d[fx] + 1;

    fa[x] = fx;

    dp[x] = d[x] % 2 ? 1 : -1;

    for (int y : G[x]) {

        if (y == fx) continue;

        if (d[y]) {

            delv = y;

            delu = x;

            continue;

        }

        dfs(y, x);

        dp[x] += dp[y];

    }

}

void SolveTree() {

    if (dp[1]) {

        puts("-1");

        return;

    }

    int ans = 0;

    for (int i = 1; i <= n; i++) ans += abs(dp[i]);

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

}

void SolveOdd() {

    if (dp[1] % 2) {

        puts("-1");

        return;

    }

    int ans = 0, t = -dp[1] / 2;

    ans = abs(t);

    for (; delu; delu = fa[delu]) dp[delu] += t;

    for (; delv; delv = fa[delv]) dp[delv] += t;

    for (int i = 1; i <= n; i++) ans += abs(dp[i]);

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

}

void SolveEven() {

    if (dp[1]) {

        puts("-1");

        return;

    }

    int ans, top = 0, mid;

    for (; delv != delu; delv = fa[delv]) {

        q[++top] = dp[delv];

        d[delv] = 0;

    }

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

    mid = q[(top + 1) / 2];

    ans = abs(mid);

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

        if (d[i]) ans += abs(dp[i]);

    for (int i = 1; i <= top; i++) ans += abs(q[i] - mid);

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

}

int main() {

    scanf("%d%d", &n, &m);

    for (int i = 1; i <= m; i++) {

        scanf("%d%d", &x, &y);

        G[x].push_back(y);

        G[y].push_back(x);

    }

    dfs(1, 0);

    if (m == n - 1)

        SolveTree();

    else if ((d[delv] - d[delu]) % 2)

        SolveEven();

    else

        SolveOdd(); 

}

// 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: