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;
template <typename T, typename U>
istream &operator>>(istream &in, pair<T, U> &data) {
  in >> data.first >> data.second;
  return in;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &vect) {
  for (unsigned i = 0; i < vect.size(); i++) {
    in >> vect[i];
  }
  return in;
}
namespace std {
template <>
struct hash<std::pair<int, int>> {
  inline size_t operator()(const std::pair<int, int> &v) const {
    std::hash<int> int_hasher;
    return int_hasher(v.first) ^ int_hasher(v.second);
  }
};
}  // namespace std
template <typename Out>
void Split(const std::string &s, char delim, Out result) {
  std::stringstream ss(s);
  std::string item;
  while (std::getline(ss, item, delim)) {
    *(result++) = item;
  }
}
std::vector<std::string> str_split(const std::string &s, char delim) {
  std::vector<std::string> elems;
  Split(s, delim, std::back_inserter(elems));
  return elems;
}
class MinSegmentTree {
 public:
  int _n;
  vector<int> _arr;
  MinSegmentTree(vector<int> &arr) : _n(arr.size()), _arr(2 * _n) {
    for (int i = 0; i < _n; i++) {
      _arr[i + _n] = arr[i];
    }
    for (int i = _n - 1; i > 0; i--) {
      _arr[i] = min(_arr[i << 1], _arr[(i << 1) + 1]);
    }
  }
  int query(int l, int r) {
    if (r < l) swap(r, l);
    int currmin = _arr[_n + l];
    for (l += _n, r += _n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) currmin = min(currmin, _arr[l++]);
      if (r & 1) currmin = min(currmin, _arr[--r]);
    }
    return currmin;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  long long n, q;
  cin >> n >> q;
  vector<bool> a(n);
  vector<long long> sa_kec(n + 2);
  vector<long long> cisto(n + 2);
  sa_kec[0] = 0;
  cisto[0] = 1;
  for (int i = 1; i < n + 2; i++) {
    sa_kec[i] = (sa_kec[i - 1] * 2 + 1) % 1000000007;
    cisto[i] = (cisto[i - 1] * 2) % 1000000007;
  }
  string inp;
  cin >> inp;
  for (int i = 0; i < n; i++) {
    a[i] = (inp[i] == '1');
  }
  vector<long long> prefix_ones(n + 1);
  vector<long long> prefix_zeros(n + 1);
  prefix_ones[0] = prefix_zeros[0] = 0;
  for (int i = 0; i < n; i++) {
    prefix_ones[i + 1] = prefix_ones[i];
    prefix_zeros[i + 1] = prefix_zeros[i];
    if (a[i]) {
      prefix_ones[i + 1]++;
    } else {
      prefix_zeros[i + 1]++;
    }
  }
  for (int qi = 0; qi < q; qi++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    long long num_ones = prefix_ones[r + 1] - prefix_ones[l];
    long long num_zero = prefix_zeros[r + 1] - prefix_zeros[l];
    long long sol = sa_kec[num_ones] * cisto[num_zero];
    sol %= 1000000007;
    cout << sol << endl;
  }
  return 0;
}