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;
long long fenwick[100005];
void update(long long value, long long where, long long n) {
  while (where <= n) {
    fenwick[where] += value;
    where += (where & -where);
  }
}
long long query(long long where) {
  long long res = 0;
  while (where) {
    res += fenwick[where];
    where -= (where & -where);
  }
  return res;
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long p[100007];
long long exp(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long half = exp(a, b / 2) % 1000000007;
  if (b % 2 == 1)
    return ((half * half) % 1000000007 * a % 1000000007) % 1000000007;
  return (half * half) % 1000000007;
}
int main() {
  fastio();
  long long n, q;
  cin >> n >> q;
  vector<long long> arr;
  memset(fenwick, 0, sizeof fenwick);
  memset(p, 0, sizeof p);
  p[0] = 0;
  for (int i = 1; i < 100007; i++) {
    if (i == 1) {
      p[i] = 2;
      continue;
    }
    long long lo = exp(2, i);
    p[i] = ((p[i - 1] % 1000000007) + lo % 1000000007) % 1000000007;
    p[i] %= 1000000007;
  }
  string s;
  cin >> s;
  for (long long i = 0; i < n; i++) update((long long)s[i] - '0', i + 1, n);
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long x = query(r) - query(l - 1);
    long long ans = 0;
    if (r - l + 1 == x)
      cout << ((((p[(r - l)] % 1000000007 -
                  p[(r - l + 1 - x) - 1] % 1000000007) +
                 1000000007) %
                1000000007) +
               1) %
                  1000000007
           << "\n";
    else
      cout << (((p[(r - l)] % 1000000007 -
                 p[(r - l + 1 - x) - 1] % 1000000007) +
                1000000007) %
               1000000007)
           << "\n";
  }
  return 0;
}