Today Adilbek is taking his probability theory test. Unfortunately, when Adilbek arrived at the university, there had already been a long queue of students wanting to take the same test. Adilbek has estimated that he will be able to start the test only T seconds after coming. 

Fortunately, Adilbek can spend time without revising any boring theorems or formulas. He has an app on this smartphone which contains n Japanese crosswords to solve. Adilbek has decided to solve them all one by one in the order they are listed in the app, without skipping any crossword. For each crossword, a number t_i is given that represents the time it takes an average crossword expert to solve this crossword (the time is given in seconds).

Adilbek is a true crossword expert, but, unfortunately, he is sometimes unlucky in choosing the way to solve the crossword. So, it takes him either t_i seconds or t_i + 1 seconds to solve the i-th crossword, equiprobably (with probability 1/2 he solves the crossword in exactly t_i seconds, and with probability 1/2 he has to spend an additional second to finish the crossword). All these events are independent.

After T seconds pass (or after solving the last crossword, if he manages to do it in less than T seconds), Adilbek closes the app (if he finishes some crossword at the same moment, that crossword is considered solved; otherwise Adilbek does not finish solving the current crossword at all). He thinks it would be an interesting probability theory problem to calculate E — the expected number of crosswords he will be able to solve completely. Can you calculate it? 

Recall that the expected value of a discrete random variable is the probability-weighted average of all possible values — in this problem it means that the expected value of the number of solved crosswords can be calculated as E = ∑ _{i = 0}^{n} i p_i, where p_i is the probability that Adilbek will solve exactly i crosswords. 

We can represent E as rational fraction P/Q with Q > 0. To give the answer, you should print P ⋅ Q^{-1} mod (10^9 + 7).

Input

The first line contains two integers n and T (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ T ≤ 2 ⋅ 10^{14}) — the number of crosswords and the time Adilbek has to spend, respectively.

The second line contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ 10^9), where t_i is the time it takes a crossword expert to solve the i-th crossword.

Note that Adilbek solves the crosswords in the order they are given in the input without skipping any of them.

Output

Print one integer — the expected value of the number of crosswords Adilbek solves in T seconds, expressed in the form of P ⋅ Q^{-1} mod (10^9 + 7).

Examples

Input


3 5
2 2 2


Output


750000007


Input


3 5
2 1 2


Output


125000003

Note

The answer for the first sample is equal to 14/8.

The answer for the second sample is equal to 17/8.
//package educational.round68;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class F {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni();
		long T = nl();
		int[] a = na(n);
		long[] high = new long[n+1];
		for(int i = 0;i <= n;i++){
			high[i] = T;
			if(i < n){
				T -= a[i];
			}
		}
		// C(i, [T-i, T])
		long[] H = go(high);
		long m = 1;
		long E = 0;
		long pnum = 0;
		int mod = 1000000007;
		for(int i = n;i >= 0;i--){
			long num = (H[i]) * m % mod;
			E += (num-pnum) % mod * i % mod;
			pnum = num;
			m = m * 2 % mod;
		}
		E %= mod;
		if(E < 0)E += mod;
		out.println(E* pow(invl(2, mod), n, mod) % mod);
	}
	
	// sum_{j<=h[i]} C(i, j)
	// h[i] is decremental
	long[] go(long[] h)
	{
		long s = 1;
		long[] ret = new long[h.length];
		int tail = 0;
		for(int i = 0;i < h.length;i++){
			if(h[i] < 0)break;
			
			if(i > 0){
				s = s * 2;
				s += C(i-1, tail+1, mod, fif);
				s %= mod;
				tail++;
				
				while(tail > h[i]){
					s -= C(i, tail, mod, fif);
					tail--;
				}
				s %= mod;
				if(s < 0)s += mod;
			}
			
			ret[i] = s;
		}
		return ret;
	}
	
	int mod = 1000000007;
	int[][] fif = enumFIF(400005, mod);
	
	public static long C(int n, int r, int mod, int[][] fif) {
		if (n < 0 || r < 0 || r > n)
			return 0;
		return (long) fif[0][n] * fif[1][r] % mod * fif[1][n - r] % mod;
	}
	
	public static int[][] enumFIF(int n, int mod) {
		int[] f = new int[n + 1];
		int[] invf = new int[n + 1];
		f[0] = 1;
		for (int i = 1; i <= n; i++) {
			f[i] = (int) ((long) f[i - 1] * i % mod);
		}
		long a = f[n];
		long b = mod;
		long p = 1, q = 0;
		while (b > 0) {
			long c = a / b;
			long d;
			d = a;
			a = b;
			b = d % b;
			d = p;
			p = q;
			q = d - c * q;
		}
		invf[n] = (int) (p < 0 ? p + mod : p);
		for (int i = n - 1; i >= 0; i--) {
			invf[i] = (int) ((long) invf[i + 1] * (i + 1) % mod);
		}
		return new int[][] { f, invf };
	}


	
	public static long pow(long a, long n, long mod) {
		//		a %= mod;
		long ret = 1;
		int x = 63 - Long.numberOfLeadingZeros(n);
		for (; x >= 0; x--) {
			ret = ret * ret % mod;
			if (n << 63 - x < 0)
				ret = ret * a % mod;
		}
		return ret;
	}

	
	public static long invl(long a, long mod) {
		long b = mod;
		long p = 1, q = 0;
		while (b > 0) {
			long c = a / b;
			long d;
			d = a;
			a = b;
			b = d % b;
			d = p;
			p = q;
			q = d - c * q;
		}
		return p < 0 ? p + mod : p;
	}

	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new F().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}
