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;
int delta[4][2] = {{-1, -0}, {0, 1}, {1, 0}, {0, -1}};
long long calc_pow(long long a, long long p, long long m = 1000000007) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * a) % m;
    a = (a * a) % m;
    p >>= 1;
  }
  return res;
}
long long calc_pow_without_mod(long long a, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = res * a;
    a = a * a;
    p >>= 1;
  }
  return res;
}
void func() {
  long long n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  char arr[n + 1];
  for (int i = 0; i < n; i++) arr[i + 1] = s[i];
  long long pre[n + 1];
  for (int i = 0; i <= n; i++) pre[i] = 0;
  for (int i = 1; i <= (n); i++) {
    pre[i] = pre[i - 1];
    if (arr[i] == '1') pre[i]++;
  }
  for (int i = 1; i <= (q); i++) {
    int l, r;
    cin >> l >> r;
    long long ones = pre[r] - pre[l - 1];
    long long zeroes = r - l + 1 - ones;
    long long val = (calc_pow(2, ones) - 1 + 1000000007) % 1000000007;
    long long new_val = (calc_pow(2, zeroes) - 1 + 1000000007) % 1000000007;
    new_val = (new_val * val) % 1000000007;
    cout << (val + new_val) % 1000000007 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  func();
  return 0;
}