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.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ii = pair<int, int>;
constexpr int MAXN = 5 + 100000;
constexpr int MOD = 7 + 1000000000;
int oac[MAXN];
int add(int a, int b) {
  int ans = a + b;
  if (ans >= MOD) ans -= MOD;
  return ans;
}
int sub(int a, int b) {
  int ans = a - b;
  if (ans < 0) ans += MOD;
  return ans;
}
int mul(int a, int b) { return (int)((1LL * a * b) % MOD); }
int fpow(int b, int e) {
  int ans = 1, p = b;
  for (; e; e >>= 1) {
    if (e & 1) ans = mul(ans, p);
    p = mul(p, p);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  string s;
  int n, q;
  cin >> n >> q >> s;
  for (int i = (int)0; i < (int)n; ++i) {
    oac[i] = (s[i] == '1');
    if (i >= 1) oac[i] += oac[i - 1];
  }
  while (q--) {
    int first, second;
    cin >> first >> second;
    --first;
    --second;
    int o = oac[second];
    if (first >= 1) o -= oac[first - 1];
    if (o == 0) {
      cout << 0 << '\n';
    } else {
      int sum1 = sub(fpow(2, o), 1);
      int sum2 = mul(sum1, sub(fpow(2, 1 + second - first - o), 1));
      cout << add(sum1, sum2) << '\n';
    }
  }
  return 0;
}
