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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long n1, long long n2) {
  if (!n1) return n2;
  if (!n2) return n1;
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  base %= 1000000007;
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 1000000007;
    base = (base * base) % 1000000007;
    exponent /= 2;
  }
  ans %= 1000000007;
  return ans;
}
int arr[1000100 + 1];
long long dp[1000100 + 1];
int main() {
  clock_t begin = clock();
  ;
  int i, j, k, n, q, l, r;
  scanf("%d", &n);
  scanf("%d", &q);
  string s;
  cin >> s;
  for (i = 1; i <= n; i++) {
    arr[i] = arr[i - 1];
    if (s[i - 1] == '1') arr[i]++;
  }
  while (q--) {
    scanf("%d", &l);
    scanf("%d", &r);
    int ones = arr[r] - arr[l - 1];
    int zeroes = ((r - l + 1) - ones);
    long long answer = powmod(2, ones) - 1;
    answer = (answer * powmod(2, zeroes)) % 1000000007;
    printf("%lld\n", answer);
  }
  clock_t end = clock();
  double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
  fprintf(stderr, "\nTime elapsed : %.3f seconds\n", elapsed_secs);
  return 0;
  ;
}