Problem: The gym leaders were fascinated by the evolutions which took place at Felicity camp. So, they were curious to know about the secret behind evolving Pokemon. 

The organizers of the camp gave the gym leaders a PokeBlock, a sequence of n ingredients. Each ingredient can be of type 0 or 1. Now the organizers told the gym leaders that to evolve a Pokemon of type k (k ≥ 2), they need to make a valid set of k cuts on the PokeBlock to get smaller blocks.

Suppose the given PokeBlock sequence is b0b1b2... bn - 1. You have a choice of making cuts at n + 1 places, i.e., Before b0, between b0 and b1, between b1 and b2, ..., between bn - 2 and bn - 1, and after bn - 1.

The n + 1 choices of making cuts are as follows (where a | denotes a possible cut):

| b0 | b1 | b2 | ... | bn - 2 | bn - 1 |

Consider a sequence of k cuts. Now each pair of consecutive cuts will contain a binary string between them, formed from the ingredient types. The ingredients before the first cut and after the last cut are wasted, which is to say they are not considered. So there will be exactly k - 1 such binary substrings. Every substring can be read as a binary number. Let m be the maximum number out of the obtained numbers. If all the obtained numbers are positive and the set of the obtained numbers contains all integers from 1 to m, then this set of cuts is said to be a valid set of cuts.

For example, suppose the given PokeBlock sequence is 101101001110 and we made 5 cuts in the following way:

10 | 11 | 010 | 01 | 1 | 10

So the 4 binary substrings obtained are: 11, 010, 01 and 1, which correspond to the numbers 3, 2, 1 and 1 respectively. Here m = 3, as it is the maximum value among the obtained numbers. And all the obtained numbers are positive and we have obtained all integers from 1 to m. Hence this set of cuts is a valid set of 5 cuts.

A Pokemon of type k will evolve only if the PokeBlock is cut using a valid set of k cuts. There can be many valid sets of the same size. Two valid sets of k cuts are considered different if there is a cut in one set which is not there in the other set.

Let f(k) denote the number of valid sets of k cuts. Find the value of <image>. Since the value of s can be very large, output s modulo 109 + 7.

Input

The input consists of two lines. The first line consists an integer n (1 ≤ n ≤ 75) — the length of the PokeBlock. The next line contains the PokeBlock, a binary string of length n.

Output

Output a single integer, containing the answer to the problem, i.e., the value of s modulo 109 + 7.

Examples

Input

4
1011


Output

10


Input

2
10


Output

1

Note

In the first sample, the sets of valid cuts are:

Size 2: |1|011, 1|01|1, 10|1|1, 101|1|.

Size 3: |1|01|1, |10|1|1, 10|1|1|, 1|01|1|.

Size 4: |10|1|1|, |1|01|1|.

Hence, f(2) = 4, f(3) = 4 and f(4) = 2. So, the value of s = 10.

In the second sample, the set of valid cuts is:

Size 2: |1|0.

Hence, f(2) = 1 and f(3) = 0. So, the value of s = 1.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;


public class Main {
	static InputReader in = new InputReader(System.in);
	public static void main(String args[]) {
		int dp[][]=new int[76][1<<20];
		int n = in.readInt();
		String s=in.readString();
		for(int i=0;i<n;i++) {
			int b=0;
			for(int j=i;j>=0;j--) {
				if(s.charAt(j)=='1' && (i-j)>5) break;
				if(s.charAt(j)=='1') b+=1<<(i-j);
				if(b>20) break;
				if(b>0) dp[i][1<<(b-1)]++;
			}
		}
		for(int i=0;i<n;i++) {
			int b=0;
			for(int j=i;j>0;j--) {
				if(s.charAt(j)=='1' && (i-j)>5) break;
				if(s.charAt(j)=='1') b+=1<<(i-j);
				if(b>20) break;
				if(b==0) continue;
				for(int k=0;k<1<<20;k++) {
					int o=1<<(b-1);
					dp[i][k|o] += dp[j-1][k];
					if(dp[i][k|o]>=1000000007)
						dp[i][k|o]-=1000000007;
				}
			}
		}
		int ans=0;
		for(int i=0;i<n;i++) {
			for(int k=0;k<1<<20;k++) {
				if(((k+1)&k)==0) {
					ans+=dp[i][k];
					if(ans>=1000000007) ans-=1000000007;
				}
			}
		}
		System.out.println(ans);
	}
}
	/*
class dir{
	boolean l,r,u,d;
}
class Node extends ArrayList<Node>{

	private static final long serialVersionUID = 1L;
	double x,y;
	int i;
	boolean visited=false;
	Node(int x,int y,int i){
		this.x = x;
		this.y = y;
		this.i=i;
	}
	void reset(){
		visited = false;
	}
	boolean dfs(dir d,double mid,double w,double h,double [][]dists,double cmpr){
		visited = true;
		if(x+mid>=w)
			d.r=true;
		if(x-mid<=0.0)
			d.l=true;
		if(y+mid>=h)
			d.u=true;
		if(y-mid<=0.0)
			d.d=true;
		if((d.l && d.d) || (d.u && d.r) 
				|| (d.l && d.r) || (d.u && d.d)) {
			return true;
		}
		for(Node n : this) {
			if(!n.visited) {
				boolean fl=false;
				if(dists[i][n.i]<=cmpr)
					fl|=n.dfs(d,mid,w,h,dists,cmpr);
				if(fl) return fl;
			}
		}
		return false;
	}
	double dist(Node n1,Node n2) {
		return sq(n1.x-n2.x)+sq(n1.y-n2.y);
	}
	double sq(double a) {
		return a*a;
	}
}
class ChristmasBinaryTree{
    public int count(long d, int[] left, int[] right) {
    	int n = left.length;
    	int dp[][][]=new int[120][120][64];
    	for(int i=0;i<n;i++) {
    		dp[i][left[i]][0]=1;
    		dp[i][right[i]][0]=1;
    	}
    	while(d>0) {
    		long k=0;
    		for(;1<<k<=d;k++) {
    			for(int i=0;i<n;i++) {
    				for(int j=0;j<n;j++) {
    					for(int l=0;l<n;l++) {
    						dp[i][l][(int)k+1]+=
    								dp[i][j][(int)k]*dp[j][l][(int)k];
    						
    					}
    				}
    			}
    		}
        	int ndp[][][]=new int[120][120][64];
    		for(int i=0;i<n;i++) {
    			for(int j=0;j<n;j++) {
    				ndp[i][j][0]=dp[i][j][(int)k];
    			}
    		}
    		dp=ndp;
    		d-=1<<(k-1);
    	}
    	int cn=0;
    	for(int i=0;i<n;i++) {
    		for(int j=0;j<n;j++) {
    			cn+=dp[i][j][0]*dp[i][j][0];
    		}
    	}
    	return cn;
    }
}
/*
public class Main {
	private static InputReader in = new InputReader(System.in);
	public static void main(String args[]) {
		StringBuilder s = new StringBuilder();
		int t = in.readInt();
		while(t>0) {
			
			t--;
		}
		System.out.println(s);
	}

	/*public static int solve(String s, int i,int j,
    		int dp[][],boolean visited[][]) {
    	if(i>j) return 0;
    	if(i==j) return dp[i][j]=1;
    	if(visited[i][j]) return dp[i][j];
    	visited[i][j]=true;
    	solve(s,i+1,j,dp,visited);
    	solve(s,i,j-1,dp,visited);
    	if(s.charAt(i)==s.charAt(j)) {
    		if(i+1==j)
    			return dp[i][j]=2;
    		else {
    			return dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+1;
    		}
    	}else {
    		if(i+1==j) return dp[i][j]=1;
    		return dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1];
    	}
    }*/
	/*
    public static void main(String args[]) {
    	int t = in.nextInt();
    	while(t>0) {
    		int m = in.nextInt(), n = in.nextInt();
    		long p = in.nextLong();
    		long dp[][][]=new long[11][300][300];

    		for(int i=1;i<=m;i++)
    			dp[1][i][i]=1;
    		for(int i=1;i<n;i++) {
    			for(int j=1;j<m;j++) {
    				for(int k=1;k<=j;k++) {
    					for(int l=k;l>=1;l--) {
    						if(j+l<=m)
    							dp[i+1][j+l][l]+=dp[i][j][k];
    					}
    				}
    			}
    		}
    		System.out.println(dp[n][m-1][2]);
    		for(int i=n;i>0;i--) {
    			for(int j=1;j<=m;j++) {
    				if(dp[i][m-j][j]>=p) {
    					//System.out.print(j+" ");
    					m-=j;
    					break;
    				}else {
    					p-=dp[i][m-j][j];
    					System.out.print(p+" ");
    				}
    			}
    		}
    		System.out.println();
    		t--;
    	}

    }*/

class sh implements Comparable<sh>{
	int h,w,i;
	sh(int a,int b,int c){
		w=a;
		h=b;
		i=c;
	}
	@Override
	public int compareTo(sh o) {
		if(w==o.w) return h-o.h;
		return w-o.w;
	}

}

class InputReader {

	private InputStream stream;
	private byte[] buf = new byte[1024];
	private int curChar;
	private int numChars;
	private SpaceCharFilter filter;

	public InputReader(InputStream stream) {
		this.stream = stream;
	}

	public byte read() {
		if (numChars == -1)
			throw new InputMismatchException();
		if (curChar >= numChars) {
			curChar = 0;
			try {
				numChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (numChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}

	public int readInt() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public String readString() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	}

	public boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}

	public String next() {
		return readString();
	}

	public interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}

class IOUtils {

	public static int[] readIntArray(InputReader in, int size) {
		int[] array = new int[size];
		for (int i = 0; i < size; i++)
			array[i] = in.readInt();
		return array;
	}
}