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;
bool cmp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.first == b.first) return (a.second < b.second);
  return a.first < b.first;
}
long long nCr(long long n, long long r) {
  long long res = 1;
  for (int i = n; i > max(r, n - r); i--) res *= i, res %= 1000000007;
  for (int i = 2; i <= min(r, n - r); i++) res /= i;
  return res % 1000000007;
}
long long exp(long long n, long long j) {
  long long i = n, res = 1;
  while (j > 0) {
    if (j & 1) res *= i;
    res %= 1000000007;
    i = i * i;
    i %= 1000000007;
    j >>= 1;
  }
  return res % 1000000007;
}
int main() {
  int T = 1;
  while (T--) {
    long long n, q, i, j;
    cin >> n >> q;
    string a;
    cin >> a;
    long long cone[n + 1];
    cone[0] = 0;
    for (i = 1; i < n + 1; i++) {
      cone[i] = (a[i - 1] == '1') + cone[i - 1];
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      long long ones = cone[r] - cone[l - 1];
      long long zeros = (r - l + 1) - ones;
      long long x = (exp(2ll, ones) - 1ll) % 1000000007;
      long long y = (x * (exp(2ll, zeros) - 1)) % 1000000007;
      cout << ((x + y) % 1000000007) << '\n';
    }
  }
}