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 MOD = 1e9 + 7;
class segTree {
 private:
  vector<int> seg;

 public:
  segTree(int n = 0) { seg.assign(4 * n + 7, 0); }
  void build(int k, int l, int r, int x, int val) {
    if (x < l || r < x) return;
    if (l == r) {
      seg[k] = val;
      return;
    }
    int mid = (l + r) / 2;
    build(k * 2, l, mid, x, val);
    build(k * 2 + 1, mid + 1, r, x, val);
    seg[k] = seg[k * 2] + seg[k * 2 + 1];
  }
  int get(int k, int l, int r, int u, int v) {
    if (v < l || u > r) return 0;
    if (u <= l && r <= v) return seg[k];
    int mid = (l + r) / 2;
    return get(k * 2, l, mid, u, v) + get(k * 2 + 1, mid + 1, r, u, v);
  }
};
int p[200000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  char s;
  segTree seg(n);
  for (int i = 1; i <= n; i++) {
    cin >> s;
    seg.build(1, 1, n, i, s - '0');
  }
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = (1ll * p[i - 1] * 2) % MOD;
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    int g1 = seg.get(1, 1, n, u, v);
    int g0 = v - u + 1 - g1;
    long long ans = (p[g1] - 1 + MOD) % MOD;
    ans = 1ll * p[g0] * ans % MOD;
    cout << ans << "\n";
  }
  return 0;
}