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 PYTHON):

Solution:
from __future__ import division
from sys import stdin, stdout
# from fractions import gcd
# from math import *
# from operator import mul
# from functools import reduce
# from copy import copy
from collections import deque, defaultdict, Counter

rstr = lambda: stdin.readline().strip()
rstrs = lambda: [str(x) for x in stdin.readline().split()]
rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]
rstr_2d = lambda n: [rstr() for _ in range(n)]
rint_2d = lambda n: [rint() for _ in range(n)]
rints_2d = lambda n: [rints() for _ in range(n)]
pr = lambda args, sep: stdout.write(sep.join(map(str, args)) + '\n')
out = []
add = lambda a, b: (a + b) % mod


def arr_sum():
    tem0, tem1 = [0], [0]
    for i in range(n):
        tem0.append(tem0[i] + int(s[i] == '0'))
        tem1.append(tem1[i] + int(s[i] == '1'))
    return tem0, tem1


n, q = rints()
s, qur, mod = rstr(), rints_2d(q), 10 ** 9 + 7
zero, one = arr_sum()

for l, r in qur:
    all0, all1 = zero[r] - zero[l - 1], one[r] - one[l - 1]
    if not all1:
        out.append(0)
    else:
        v1 = add(-1, add(-pow(2, all0, mod), 1))
        out.append(add(pow(2, all1 + all0, mod), v1))

pr(out, '\n')