// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 2.219168337960785x):
#include<bits/stdc++.h>

using namespace std;

struct edge{int to,nxt;}e[200005];

int n,m,et,rx,rt,head[100005],fa[100005],a[100005],tot,b[100005],r;

inline void adde(int x,int y) {e[++tot]=(edge){y,head[x]},head[x]=tot;}

inline int getf(int x) {return fa[x]==x?x:fa[x]=getf(fa[x]);}

inline void dfs(int x,int fl,int f=0)

{

	a[x]=fl,fa[x]=f;

	for(int i=head[x];i;i=e[i].nxt) if(e[i].to!=f) dfs(e[i].to,-fl,x),a[x]+=a[e[i].to];

}

int main()

{

	scanf("%d%d",&n,&m),et=0,rt=1,memset(head,0,sizeof(head));

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

	for(int i=1,x,y;i<=m;i++) {scanf("%d%d",&x,&y);if(getf(x)!=getf(y)) adde(x,y),adde(y,x),fa[getf(x)]=getf(y);else rt=x,rx=y;}

	memset(a,0,sizeof(a)),fa[rt]=0,dfs(rt,1),tot=0,r=0;

	if(n!=m) {if(a[rt]) return puts("-1"),0;}

	else

	{

		for(int x=rx;x;x=fa[x]) b[++tot]=a[x];

		if(tot&1)

		{

			if(a[rt]&1) return puts("-1"),0;

			for(int x=rx;x;x=fa[x]) a[x]-=(a[rt]>>1);

		}

		else

		{

			if(a[rt]) return puts("-1"),0;

			sort(b+1,b+tot+1);

			for(int x=rx;x;x=fa[x]) a[x]-=b[tot>>1];

		}

	}

	for(int i=1;i<=n;i++) r+=abs(a[i]);

	return printf("%d\n",r),0;

}