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;
using namespace std;
string s;
long long tree_bit[200100];
int lim;
void update_bit(int idx, int val) {
  while (idx <= lim) {
    tree_bit[idx] += val;
    if (tree_bit[idx] >= 1000000007) tree_bit[idx] %= 1000000007;
    idx += (idx & -idx);
  }
}
long long query_bit(int idx) {
  if (!idx) return 0;
  long long sum = 0;
  while (idx != 0) {
    sum += tree_bit[idx];
    idx -= (idx & -idx);
    if (sum >= 1000000007) sum %= 1000000007;
  }
  return sum;
}
long long pw2[100100];
long long get_progression(long long a, long long pod) {
  if (a >= 1000000007) a %= 1000000007;
  long long ret = a * (pw2[pod] - 1);
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  pw2[0] = 1;
  for (int i = 1; i <= 100010; i++) {
    pw2[i] = (pw2[i - 1] << 1);
    if (pw2[i] >= 1000000007) pw2[i] %= 1000000007;
  }
  cin >> n >> q;
  cin >> s;
  lim = n;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      update_bit(i + 1, 1);
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    int range = r - l;
    int ones = query_bit(r) - query_bit(l);
    long long ans = 0;
    ans = get_progression(1, ones);
    ans += get_progression(ans, range - ones);
    if (ans >= 1000000007) ans %= 1000000007;
    cout << ans << endl;
  }
  return 0;
}