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>
ostream& operator<<(ostream& os, vector<T>& v) {
  if (v.size() == 0) {
    os << "empty vector\n";
    return os;
  }
  for (auto element : v) os << element << " ";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, pair<T, second>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& v) {
  if (v.size() == 0) {
    os << "empty set\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& v) {
  if (v.size() == 0) {
    os << "empty multiset\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, map<T, second>& v) {
  if (v.size() == 0) {
    os << "empty map\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "{";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << "(" << (*it).first << " : " << (*it).second << ")";
    if (it != endit) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<vector<T>>& v) {
  for (auto& subv : v) {
    for (auto& e : subv) os << e << " ";
    os << "\n";
  }
  return os;
}
bool do_debug = false;
vector<long long> fact(200001), ifact(200001);
long long power(long long x, long long n) {
  if (!n) return 1;
  if (n % 2)
    return (x * (power((x * x) % 1000000007, n / 2) % 1000000007)) % 1000000007;
  return power((x * x) % 1000000007, n / 2) % 1000000007;
}
long long inverse(long long n) { return power(n, 1000000007 - 2) % 1000000007; }
void pre_factorials() {
  fact[0] = 1;
  for (long long i = 1; i < 200001; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
  for (long long i = 0; i < 200001; i++) ifact[i] = inverse(i);
}
long long ncr(long long n, long long r) {
  if (n < r)
    return 0;
  else if (n == r)
    return 1;
  else if (r == 0)
    return 0;
  long long ans = fact[n];
  ans = (ans * ifact[r]) % 1000000007;
  ans = (ans * ifact[n - r]) % 1000000007;
  return ans;
}
void Runtime_Terror() {
  long long n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) v[i] = s[i] - '0';
  vector<pair<long long, long long>> cnt(n);
  if (v[0] == 0)
    cnt[0] = {1, 0};
  else
    cnt[0] = {0, 1};
  for (long long i = 1; i < n; i++) {
    if (v[i] == 0)
      cnt[i] = {cnt[i - 1].first + 1, cnt[i - 1].second};
    else
      cnt[i] = {cnt[i - 1].first, cnt[i - 1].second + 1};
  }
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    if (l == 1) {
      long long one = cnt[r - 1].second;
      long long zero = cnt[r - 1].first;
      long long ans = 0;
      long long tem =
          ((power(2, one) - 1) % 1000000007 + 1000000007) % 1000000007;
      ans += tem;
      long long dtem = tem;
      tem *= ((power(2, zero) - 1) % 1000000007 + 1000000007) % 1000000007;
      tem %= 1000000007;
      ans += tem;
      ans %= 1000000007;
      cout << ans << endl;
    } else {
      long long one = cnt[r - 1].second - cnt[l - 2].second;
      long long zero = cnt[r - 1].first - cnt[l - 2].first;
      long long ans = 0;
      long long tem =
          ((power(2, one) - 1) % 1000000007 + 1000000007) % 1000000007;
      ans += tem;
      long long dtem = tem;
      tem *= ((power(2, zero) - 1) % 1000000007 + 1000000007) % 1000000007;
      tem %= 1000000007;
      ans += tem;
      ans %= 1000000007;
      cout << ans << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (long long i = 0; i < t; i++) Runtime_Terror();
  return 0;
}