Problem: Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.

Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7). 

Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.

Input

The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).

The second line contains n space-separated integers f1, f2, ... fn (0 ≤ fi ≤ 1012).

Output

Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).

Examples

Input

2 3
1 3


Output

2


Input

2 4
2 2


Output

1


Input

3 5
1 3 2


Output

3

Note

Sample 1. There are two ways of selecting 3 flowers: {1, 2} and {0, 3}.

Sample 2. There is only one way of selecting 4 flowers: {2, 2}.

Sample 3. There are three ways of selecting 5 flowers: {1, 2, 2}, {0, 3, 2}, and {1, 3, 1}.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Task451E {

	private static final long MOD = 1000000007;
	
	private FastScanner in;
	private PrintWriter out;

	private long inverseNMinusOneFactorial;
	private long nMinusOne;

	public Task451E(InputStream in, OutputStream out) {
		this.in = new FastScanner(in);
		this.out = new PrintWriter(out);
	}

	public void solve() throws IOException {
		int n = nextInt();
		long s = nextLong();
		long f[] = new long[n];
		for (int i = 0; i < n; ++i) {
			f[i] = nextLong();
		}
		preCalc(n);
		long result = 0;
		mask:
		for (int mask = 0; mask < (1 << n); ++mask) {
			long restS = s;
			boolean subtract = false;
			for (int bit = 0; bit < n; ++bit) {
				if (((1 << bit) & mask) != 0) {
					restS -= f[bit] + 1;
					if (restS < 0) {
						continue mask;
					}
					subtract = !subtract;
				}
			}
			if (subtract) {
				result = (result + (MOD - c(restS))) % MOD;
			}
			else {
				result = (result + c(restS)) % MOD;
			}
		}
		out.println(result);
	}
	
	private void preCalc(int n) {
		long factorial = 1;
		for (int i = 1; i < n; ++i) {
			factorial = (factorial * i) % MOD;
		}
		nMinusOne = n - 1;
		inverseNMinusOneFactorial = powerMod(factorial, MOD - 2);
	}

	private long powerMod(long n, long power) {
		long result = 1;
		n = n % MOD;
		for (int bit = 1; bit <= power; bit <<= 1) {
			if ((power & bit) != 0) {
				result = (result * n) % MOD;
			}
			n = (n * n) % MOD;
		}
		return result;
	}

	public long c(long restS) {
		long result = 1;
		for (long i = restS + 1; i <= (restS + nMinusOne); ++i) {
			result = (result * (i % MOD)) % MOD;
		}
		return (result * inverseNMinusOneFactorial) % MOD;
	}

	public void run() {
		try {
			solve();
			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(InputStream stream) {
			br = new BufferedReader(new InputStreamReader(stream));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

	}

	int nextInt() {
		return Integer.parseInt(in.next());
	}

	long nextLong() {
		return Long.parseLong(in.next());
	}

	
	public static void main(String[] arg) {
		new Task451E(System.in, System.out).run();
	}
}