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 int modular_exp(long long int A, long long int B, long long int C) {
  if (B == 0) return 1;
  if (B == 1) return A;
  long long int res = A;
  if (res > C) res = res % C;
  int counter = 2;
  while (counter < B) {
    res = res * res;
    if (res > C) res = res % C;
    counter *= 2;
    if (counter >= B) break;
  }
  counter /= 2;
  return ((res % C) * modular_exp(A, B - counter, C)) % C;
}
long long int Mod(long long int A, long long int B, long long int C) {
  if (A == 0) return 0;
  if (C == 1) return 0;
  long long int res = modular_exp(A, B, C);
  if (res < 0) return C + res;
  if (B == 0) return 1;
  return res;
}
long long int ans__(long long int num_one, long long int num_zero) {
  long long int ans =
      ((Mod(2, num_one, 1000000007) - 1) + 1000000007) % 1000000007;
  long long int ans_2 =
      ((Mod(2, num_zero, 1000000007) - 1) + 1000000007) % 1000000007;
  ans_2 = (ans % 1000000007 * ans_2 % 1000000007) % 1000000007;
  return (ans % 1000000007 + ans_2 % 1000000007) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int u = 0; u < t; u++) {
    long long int n, q;
    cin >> n >> q;
    string s;
    cin >> s;
    vector<long long int> ones, zeros;
    long long int num_ones = 0;
    long long int num_zeros = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0') num_zeros++;
      if (s[i] == '1') num_ones++;
      ones.push_back(num_ones);
      zeros.push_back(num_zeros);
    }
    for (int i = 0; i < q; i++) {
      long long int l, r;
      cin >> l >> r;
      l--;
      r--;
      long long int one, zero;
      if (l > 0) {
        one = ones[r] - ones[l - 1];
        zero = zeros[r] - zeros[l - 1];
      } else if (l == 0) {
        one = ones[r];
        zero = zeros[r];
      }
      cout << ans__(one, zero) << "\n";
    }
  }
  return 0;
}