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 long long mod = 1000000007;
const long long OO = (long long)1e9;
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int dy[] = {1, 0, -1, 0, 1, -1, -1, 1};
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(4);
}
vector<int> tree;
int arr[1000000];
long long vec[100005];
void build(int s, int e, int pos) {
  if (s == e) {
    tree[pos] = arr[e];
    return;
  }
  int mid = (s + e) / 2;
  build(s, mid, pos * 2);
  build(mid + 1, e, pos * 2 + 1);
  tree[pos] = tree[pos * 2] + tree[pos * 2 + 1];
}
int getQ(int l, int r, int s, int e, int pos) {
  if (s >= l && e <= r) return tree[pos];
  if (s > r || e < l) return 0;
  int mid = (s + e) / 2;
  int x = getQ(l, r, s, mid, pos * 2);
  int y = getQ(l, r, mid + 1, e, pos * 2 + 1);
  return (x + y);
}
int main() {
  fast();
  long long power = 1;
  for (int i = 1; i <= 100000; i++) {
    vec[i] = vec[i - 1] + power;
    vec[i] %= mod;
    power *= 2;
    power %= mod;
  }
  int n, q;
  cin >> n >> q;
  char ch;
  for (int i = 0; i < n; i++) {
    cin >> ch;
    arr[i] = ch - '0';
  }
  int sz = (int)(ceil(log(n) / log(2)));
  sz = 2 * (int)(pow(2, sz));
  tree.resize(sz + 1);
  build(0, n - 1, 1);
  int l, r, u, v;
  long long ans = 0;
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    l--, r--;
    u = getQ(l, r, 0, n - 1, 1);
    v = (r - l + 1) - u;
    ans = vec[u] + ((vec[u] * vec[v]) % mod);
    ans %= mod;
    cout << ans << '\n';
  }
  return 0;
}