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;
const int MAX = 1e5 + 5;
const int BASE = 1e9 + 7;
int n, q;
bool a[MAX];
int f[MAX];
int num[MAX][2];
int Power(int x, int y) {
  if (y == 0) return 1;
  int tmp = Power(x, y / 2);
  if (y % 2) return (1LL * ((1LL * tmp * tmp) % BASE) * x) % BASE;
  return (1LL * tmp * tmp) % BASE;
}
void Preprocess() {
  f[0] = 1;
  for (int i = 1; i <= n; ++i) f[i] = (1LL * f[i - 1] + Power(2, i)) % BASE;
  for (int i = 1; i <= n; ++i) {
    num[i][0] = num[i - 1][0];
    num[i][1] = num[i - 1][1];
    if (a[i])
      ++num[i][1];
    else
      ++num[i][0];
  }
}
int Calc(int l, int r) {
  int x = num[r][0] - num[l - 1][0];
  int y = num[r][1] - num[l - 1][1];
  int M = f[y - 1];
  return (1LL * M * (1 + f[x - 1])) % BASE;
}
int main() {
  cin >> n >> q;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); ++i)
    if (s[i] == '0')
      a[i + 1] = 0;
    else
      a[i + 1] = 1;
  Preprocess();
  for (int i = 1; i <= q; ++i) {
    int l, r;
    cin >> l >> r;
    cout << Calc(l, r) << endl;
  }
  return 0;
}