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.
from bisect import bisect_right as br
from bisect import bisect_left as bl
from collections import defaultdict
import sys
import math
MAX = sys.maxsize
MOD = 10**9+7
MAXN = 10**6+10
def isprime(n):
    n = abs(int(n))
    if n < 2:
        return False
    if n == 2: 
        return True    
    if not n & 1: 
        return False
    for x in range(3, int(n**0.5) + 1, 2):
        if n % x == 0:
            return False
    return True

def mhd(a,b,x,y):
    return abs(a-x)+abs(b-y)

def numIN():
    return(map(int,sys.stdin.readline().strip().split()))

def charIN():
    return(sys.stdin.readline().strip().split())

def maxSubArraySum(a,size): 
      
    max_so_far = 0
    max_ending_here = 0
      
    for i in range(size): 
        max_ending_here = max_ending_here + a[i] 
        if max_ending_here < 0: 
            max_ending_here = 0 
        elif (max_so_far < max_ending_here): 
            max_so_far = max_ending_here 
              
    return max_so_far

def isPer(n):
	return((math.floor(n**0.5))==(math.ceil(n**0.5)))

def modInverse(a, m) : 
    m0 = m 
    y = 0
    x = 1
  
    if (m == 1) : 
        return 0
  
    while (a > 1) : 
  
        # q is quotient 
        q = a // m 
  
        t = m 
  
        # m is remainder now, process 
        # same as Euclid's algo 
        m = a % m 
        a = t 
        t = y 
  
        # Update x and y 
        y = x - q * y 
        x = t 
  
  
    # Make x positive 
    if (x < 0) : 
        x = x + m0 
  
    return x

n,q = numIN()
l = [int(i) for i in input()]
pre = []
s = 0
for i in l:
	s+=i
	pre.append(s)
for i in range(q):
	l,r = numIN()
	l-=1
	r-=1
	if l!=0:
		one = pre[r]-pre[l-1]
		zero = r-l+1-one
		x = pow(2,one,MOD)-1
		y = pow(2,zero,MOD)
		ans = x*y
		ans%=MOD
		print(ans)
	else:
		one = pre[r]
		zero = r-l+1-one
		x = pow(2,one,MOD)-1
		y = pow(2,zero,MOD)
		ans = x*y
		ans%=MOD
		print(ans)