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

public class F implements Runnable {
	public static void main (String[] args) {new Thread(null, new F(), "_cf", 1 << 28).start();}

	long MOD = (long)1e9+7;
	int MOD2 = 500000004;
	int MAX = 200200;
	long[] fact, inv, invFact;
	
	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		System.err.println("");

		fact = new long[MAX];
		inv = new long[MAX];
		inv[1] = 1; inv[0] = 1;
		for (int i = 2; i < inv.length; ++i) {
			long p1 = (MOD - (MOD/i));
			long p2 = inv[(int)(MOD%i)];
			inv[i] = p1 * p2 % MOD;
		}
		
		invFact = new long[MAX];
		fact[0] = invFact[0] = 1;
		for(int i = 1; i < MAX; i++) {
			fact[i] = fact[i-1] * i % MOD;
			invFact[i] = invFact[i-1] * inv[i] % MOD;
		}
		
		int n = fs.nextInt();
		long T = fs.nextLong();
		long[] a = new long[n];
		for(int i = 0; i < n; i++) a[i] = fs.nextLong();
		long[] lims = new long[n];
		long[] sums = new long[n];
		long[] invSums = new long[n];
		long[] pow2s = new long[n];
		
		long pref = 0;
		for(int i = 0; i < n; i++) {
			pref += a[i];
			long two = Math.max(0, T - pref);
			lims[i] = Math.min(two, i+1);
		}
		
		int curRow = 1, curCol = 0;
		long sum = 0, pow2 = 1;
		for(int i = 0; i < n; i++) {
			pow2 = mult(pow2, 2);
			pow2s[i] = pow2;
			if(i == 0) {
				while(curCol <= lims[i]) {
					sum = add(sum, nChooseK(curRow, curCol++));
				}
				sums[i] = sum;
				invSums[i] = sub(pow2, sums[i]);
				sum = add(nChooseK(curRow, curCol), mult(2, sum));
				curCol++;
				curRow++;
			}
			else {
				if(curCol <= lims[i]) {
					while(curCol <= lims[i]) {
						sum = add(sum, nChooseK(curRow, curCol++));
					}
				}
				else if(curCol-1 > lims[i]) {
					while(curCol-1 > lims[i]) {
						sum = sub(sum, nChooseK(curRow, curCol-1));
						curCol--;
					}
				}
				sums[i] = sum;
				invSums[i] = sub(pow2, sums[i]);
				sum = add(nChooseK(curRow, curCol), mult(2, sum));
				curRow++; curCol++;
			}
		}
		
		pref = 0;
		long pref2 = 0;
		for(int i = 0; i < n; i++) {
			pref += a[i]+1;
			pref2 += a[i];
			if(pref <= T) {
				sums[i] = pow2s[i]; invSums[i] = 0;
			}
			else if(pref2 > T) {
				sums[i] = 0; invSums[i] = pow2s[i];
			}
		}
		
		long[] pow2Invs = new long[n];
		pow2Invs[0] = MOD2;
		for(int i = 1; i < n; i++) {
			pow2Invs[i] = mult(pow2Invs[i-1], MOD2);
		}
		
//		System.out.println(Arrays.toString(sums));
//		System.out.println(Arrays.toString(invSums));
//		System.out.println();
		
		for(int i = 0; i < n; i++) {
			sums[i] = mult(sums[i], pow2Invs[i]);
			invSums[i] = mult(invSums[i], pow2Invs[i]);
		}
		
		long res = 0;
		for(int i = 0; i < n; i++) {
			res = add(res, sums[i]);
		}
		out.println(res);
		
//		System.out.println(Arrays.toString(sums));
//		System.out.println(Arrays.toString(invSums));
		
		out.close();
	}
	
	long nChooseK(int n, int k) {
		if(k > n) return 0;
		long num = fact[n];
		long den = invFact[k] * invFact[n-k] % MOD;
		return num * den % MOD;
	}
	
	long sub(long a, long b) {
		a -= b;
		if(a < 0) a += MOD;
		return a;
	}
	
	long mult(long a, long b) {
		return (a * b) % MOD;
	}
	
	long add(long a, long b) {
		a += b;
		a %= MOD;
		return a;
	}

	class FastScanner {
		public int BS = 1<<16;
		public char NC = (char)0;
		byte[] buf = new byte[BS];
		int bId = 0, size = 0;
		char c = NC;
		double num = 1;
		BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) {
			try {
				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
			}
			catch (Exception e) {
				in = new BufferedInputStream(System.in, BS);
			}
		}

		public char nextChar(){
			while(bId==size) {
				try {
					size = in.read(buf);
				}catch(Exception e) {
					return NC;
				}                
				if(size==-1)return NC;
				bId=0;
			}
			return (char)buf[bId++];
		}

		public int nextInt() {
			return (int)nextLong();
		}

		public long nextLong() {
			num=1;
			boolean neg = false;
			if(c==NC)c=nextChar();
			for(;(c<'0' || c>'9'); c = nextChar()) {
				if(c=='-')neg=true;
			}
			long res = 0;
			for(; c>='0' && c <='9'; c=nextChar()) {
				res = (res<<3)+(res<<1)+c-'0';
				num*=10;
			}
			return neg?-res:res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c!='.' ? cur:cur+nextLong()/num;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c>32) {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c!='\n') {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if(c>32)return true;
			while(true) {
				c=nextChar();
				if(c==NC)return false;
				else if(c>32)return true;
			}
		}
		
		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for(int i = 0; i < n; i++) res[i] = nextInt();
			return res;
		}
		
	}

}