Let's call a string good if and only if it consists of only two types of letters — 'a' and 'b' and every two consecutive letters are distinct. For example "baba" and "aba" are good strings and "abb" is a bad string.

You have a strings "a", b strings "b" and c strings "ab". You want to choose some subset of these strings and concatenate them in any arbitrarily order.

What is the length of the longest good string you can obtain this way?

Input

The first line contains three positive integers a, b, c (1 ≤ a, b, c ≤ 10^9) — the number of strings "a", "b" and "ab" respectively.

Output

Print a single number — the maximum possible length of the good string you can obtain.

Examples

Input


1 1 1


Output


4


Input


2 1 2


Output


7


Input


3 5 2


Output


11


Input


2 2 1


Output


6


Input


1000000000 1000000000 1000000000


Output


4000000000

Note

In the first example the optimal string is "baba".

In the second example the optimal string is "abababa".

In the third example the optimal string is "bababababab".

In the fourth example the optimal string is "ababab".
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int MaxN = 4e5 + 10;
const int MAXN = 120 * 1000;
void fun() {}
long long phi(long long n) {
  long long result = n;
  for (long long p = 2; p * p <= n; ++p) {
    if (n % p == 0) {
      while (n % p == 0) n /= p;
      result -= result / p;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long mult(long long a, long long b, long long mod) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    b >>= 1;
    return res;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
vector<long long> factors(long long n) {
  vector<long long> f;
  f.push_back(n);
  for (long long x = 2; x * x <= n; x++) {
    if (n % x == 0) {
      f.push_back(x);
      if ((n / x) != x) f.push_back(n / x);
    }
  }
  return f;
}
int cD(long long n) { return floor(log10(n) + 1); }
long long powmod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long inv(long long x) { return powmod(x, mod - 2); }
long long cat(long long n) {
  long long final = 1;
  for (int i = 0; i < n; i++) {
    final = final * (2 * n - i) % mod * inv(1 + i) % mod;
  }
  final = final * inv(1 + n) % mod;
  return final;
}
long long bin(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long u = bin(x, y / 2);
  if (y % 2 == 0) {
    return u * u % mod;
  } else {
    return u * u % mod * x % mod;
  }
}
long long modFact(long long n, long long mod) {
  if (n >= mod) return 0;
  long long result = 1;
  for (int i = 1; i <= n; i++) result = (result * i) % mod;
  return result;
}
double nthRoot(double n, int k) {
  return pow(k, (1.0 / k) * (log(n) / log(k)));
}
long long MAX = 1e6;
bool prime[1000005];
bool pcheck(long long a) {
  if (a == 2) return true;
  for (long long i = 2; i <= sqrt(a); i++) {
    if (a % i) continue;
    return false;
  }
  return true;
}
long long n, k, q;
long long l, t;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
std::vector<int> v[200000];
int visited[100001];
std::vector<int> v1;
void dfs(int k, int par) {
  v1.push_back(k);
  visited[k] = 1;
  for (auto x : v[k]) {
    if (visited[x] == 0) {
      dfs(x, par);
    }
  }
}
long long m;
long long sum[310000];
int main() {
  fun();
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> t >> n >> m;
  if (t == n)
    cout << t + n + m * 2;
  else
    cout << 2 * min(t, n) + 1 + (m * 2) << "\n";
  return 0;
}
