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

using namespace std;

#pragma GCC optimize("Ofast")

typedef long long lint;

typedef unsigned int uint;

typedef pair<int, int> pii;

typedef pair<lint, lint> pll;

typedef unsigned long long ulint;

#define endl '\n'

#define fst first

#define sed second

#define pb push_back

#define mp make_pair

#define rint register int

#define newline putchar('\n')

#define leave_space putchar(' ')

#define all(x) (x).begin(), (x).end()

#define reveal(x) cerr << #x << " = " << (x) << endl

#define rep(it, f, e) for (rint it = (f); it <= int(e); ++it)

#define per(it, f, e) for (rint it = (f); it >= int(e); --it)

const int MAXN = 1e5 + 10;

vector<int> edge[MAXN], cycle;

struct DSU {

	int par[MAXN];

	DSU() {

		rep (i, 1, MAXN - 1) {

			par[i] = i;

		}

	}

	inline int find(int x) {

		return par[x] == x ? x : par[x] = find(par[x]);

	}

	inline void marry(int x, int y) {

		par[find(x)] = find(y);

	}

}	Dsu;

int val[MAXN], par[MAXN];

inline void dfs(int u, int valor) {

	val[u] = valor;

	for (auto v : edge[u]) {

		if (v == par[u]) continue;

		par[v] = u;

		dfs(v, -valor);

		val[u] += val[v];

	}

}

int main() {

	ios::sync_with_stdio(false);

	cin.tie(NULL), cout.tie(NULL);

	int le1 = 1, le2 = 1;

	int n, m, u, v;

	cin >> n >> m;

	rep (i, 1, m) {

		cin >> u >> v;

		if (Dsu.find(u) != Dsu.find(v)) {

			Dsu.marry(u, v);

			edge[u].pb(v);

			edge[v].pb(u);

		} else {

			le1 = u, le2 = v;

		}

	}

	dfs(le1, 1);

	if (n == m) {

		for (int u = le2; u; u = par[u]) {

			cycle.pb(val[u]);

		}

        if (cycle.size() & 1) {

			if (val[le1] & 1) {

				cout << -1 << endl;

				return 0;

			}

			for (int u = le2; u; u = par[u]) {

				val[u] -= val[le1] / 2;

			}

        } else {

			if (val[le1]) {

				cout << -1 << endl;

				return 0;

			}

			sort(all(cycle));

			for (int u = le2; u; u = par[u]) {

				val[u] -= cycle[cycle.size() / 2 - 1];

			}

        }

	} else if (val[le1]) {

		cout << -1 << endl;

		return 0;

	}

	lint ans = 0;

	rep (i, 1, n) {

		ans += abs(val[i]);

	}

    cout << ans << endl;

	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: