The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.

Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.

Input

The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.

The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.

Output

If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.

Examples

Input

3 3
4 3 5


Output

3


Input

3 4
5 3 4


Output

2


Input

3 7
1 2 3


Output

-1

Note

In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.

In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.

In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
#include <bits/stdc++.h>
using namespace std;
map<int, int> m;
void primeFactors(int n) {
  while (n % 2 == 0) {
    m[2]++;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      m[i]++;
      n = n / i;
    }
  }
  if (n > 2) m[n]++;
}
long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int sum(long long a) {
  int sum = 0;
  while (a > 0) {
    sum = sum + (a % 10);
    a = a / 10;
  }
  return sum;
}
int count_digit(long long n) {
  int count = 0;
  while (n > 0) {
    if (n % 10 == 9) {
      count++;
      n = n / 10;
      continue;
    } else {
      return count;
      n = n / 10;
    }
  }
}
int binarySearch(int x, int y, long long z, long long v[]) {
  int low = x;
  int high = y;
  int mid = x + (y - x) / 2;
  while (low <= high) {
    if (v[mid] == z) return mid;
    if (v[mid] < z) return binarySearch(mid + 1, high, z, v);
    if (v[mid] > z) return binarySearch(low, mid - 1, z, v);
  }
  return -1;
}
long long modularExponentiation(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  else
    return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
long long binaryExponentiation(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
int binary(int n) {
  int c = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      return pow(2, c);
    }
    n = n / 2;
    c++;
  }
}
long long ceil1(long long x, long long y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
set<long long> s;
void genrate(long long n, int len, int max) {
  if (len > max) return;
  s.insert(n);
  genrate(n * 10 + 1, len + 1, max);
  genrate(n * 10 + 0, len + 1, max);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tests = 1;
  while (tests--) {
    int n;
    long long s;
    cin >> n;
    cin >> s;
    long long a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    long long sum = 0;
    for (int i = 0; i < n; i++) sum = sum + a[i];
    if (sum < s) {
      cout << -1;
      exit(0);
    } else {
      sum = 0;
      for (int i = n - 1; i > 0; i--) {
        if (s > sum + a[i] - a[0]) {
          sum = sum + a[i] - a[0];
          a[i] = a[0];
        } else if (s == sum + a[i] - a[0]) {
          cout << a[0];
          exit(0);
        } else if (s < sum + a[i] - a[0]) {
          cout << a[0];
          exit(0);
        }
      }
      if (s > sum) {
        long long x = s - sum;
        if (x % n == 0) {
          cout << a[0] - x / n;
        } else {
          a[0] = a[0] - x / n;
          cout << a[0] - 1;
        }
      }
    }
  }
}
