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.
import java.util.*;
import java.io.*;
import java.lang.*;

public class programA {
	public static void merge(long arr[], int l, int m, int r) 
    { 
        // Find sizes of two subarrays to be merged 
        int n1 = m - l + 1; 
        int n2 = r - m; 
  
        /* Create temp arrays */
        long L[] = new long [n1]; 
        long R[] = new long [n2]; 
  
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i) 
            L[i] = arr[l + i]; 
        for (int j=0; j<n2; ++j) 
            R[j] = arr[m + 1+ j]; 
  
  
        /* Merge the temp arrays */
  
        // Initial indexes of first and second subarrays 
        int i = 0, j = 0; 
  
        // Initial index of merged subarry array 
        int k = l; 
        while (i < n1 && j < n2) 
        { 
            if (L[i] <= R[j]) 
            { 
                arr[k] = L[i]; 
                i++; 
            } 
            else
            { 
                arr[k] = R[j]; 
                j++; 
            } 
            k++; 
        } 
  
        /* Copy remaining elements of L[] if any */
        while (i < n1) 
        { 
            arr[k] = L[i]; 
            i++; 
            k++; 
        } 
  
        /* Copy remaining elements of R[] if any */
        while (j < n2) 
        { 
            arr[k] = R[j]; 
            j++; 
            k++; 
        } 
    } 
  
    // Main function that sorts arr[l..r] using 
    // merge() 
   public static void sort(long arr[], int l, int r) 
    { 
        if (l < r) 
        { 
            // Find the middle point 
            int m = (l+r)/2; 
  
            // Sort first and second halves 
            sort(arr, l, m); 
            sort(arr , m+1, r); 
  
            // Merge the sorted halves 
            merge(arr, l, m, r); 
        } 
    } 
	public static void main(String[] args)throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		long s = Long.parseLong(st.nextToken());
		long arr[]  = new long[(int)n];
		st = new StringTokenizer(br.readLine());
		for(int i=0;i<n;i++) {
			arr[i] = Long.parseLong(st.nextToken());
		}
		long sum = 0;
		sort(arr, 0, n-1);
		for(int i =0;i<n;i++) {
			sum+= arr[i];
		}
		long sum2 = sum - n*(arr[0]);
		if(s>sum)System.out.println("-1");
		else if(s<=sum2)System.out.println(arr[0]);
		else {
			long rem = s-sum2;
			long t2 = (long)Math.ceil((double)rem/(double)n);
			long ans = arr[0] - t2;
			System.out.println(ans);
		}
	}

}
