Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in JAVA):

Solution:
import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.nio.charset.StandardCharsets;
import java.util.Arrays;
import java.util.InputMismatchException;

public class C1062 {
	static class Solver {

		int N, Q, cs[];
		long del[], sum[], MOD = 1_000_000_007;

		long add(long a, long b) { return (a + b) % MOD; }
		long mul(long a, long b) { return (a * b) % MOD; }
		long pow(long b, long e) {
			if (e == 0) return 1;
			if (e == 1) return b;
			long sq = pow(b, e >> 1);
			sq = mul(sq, sq);
			return (e & 1) == 0 ? sq : mul(sq, b);
		}

		void solve(int testNumber, FastScanner s, PrintWriter out) {
			N = s.nextInt();
			Q = s.nextInt();
			cs = new int[N + 1];
			char[] ch = s.next().toCharArray();
			for (int i = 1; i <= N; i++)
				cs[i] += cs[i - 1] + ch[i - 1] - '0';
			while (Q-- > 0) {
				int l = s.nextInt(), r = s.nextInt(), len = r - l + 1;
				long z = cs[r] - cs[l - 1], o = len - z;
				long oz = add(MOD - 1, pow(2, z)); // ones
				long tot = add(oz, mul(oz, add(MOD - 1, pow(2, o))));
				out.println(tot);
			}
		}

	}

	final static boolean cases = false;

	public static void main(String[] args) {

		FastScanner s = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		Solver solver = new Solver();
		for (int t = 1, T = cases ? s.nextInt() : 1; t <= T; t++)
			solver.solve(t, s, out);

		out.close();

	}

	static int min(int a, int b) {
		return a < b ? a : b;
	}

	static int max(int a, int b) {
		return a > b ? a : b;
	}

	static long min(long a, long b) {
		return a < b ? a : b;
	}

	static long max(long a, long b) {
		return a > b ? a : b;
	}

	static int swap(int a, int b) {
		return a;
	}

	static Object swap(Object a, Object b) {
		return a;
	}

	static String ts(Object... o) {
		return Arrays.deepToString(o);
	}

	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

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

		public FastScanner(File f) throws FileNotFoundException {
			this(new FileInputStream(f));
		}

		public FastScanner(String s) {
			this.stream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
		}

		int 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++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

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

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

		public double nextDouble() {
			return Double.parseDouble(next());
		}

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

		// Jacob Garbage

		public int[] nextIntArray(int N) {
			int[] ret = new int[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextInt();
			return ret;
		}

		public int[][] next2DIntArray(int N, int M) {
			int[][] ret = new int[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextIntArray(M);
			return ret;
		}

		public long[] nextLongArray(int N) {
			long[] ret = new long[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextLong();
			return ret;
		}

		public long[][] next2DLongArray(int N, int M) {
			long[][] ret = new long[N][];
			for (int i = 0; i < N; i++)
				ret[i] = nextLongArray(M);
			return ret;
		}

		public double[] nextDoubleArray(int N) {
			double[] ret = new double[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDouble();
			return ret;
		}

		public double[][] next2DDoubleArray(int N, int M) {
			double[][] ret = new double[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDoubleArray(M);
			return ret;
		}

	}
}