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 N = 1e5 + 100;
const int mod = 1e9 + 7;
inline pair<long long, long long> mp(long long a, long long b) {
  pair<long long, long long> temp;
  temp.first = a;
  temp.second = b;
  return temp;
}
long long read_int() {
  char r;
  bool start = false, neg = false;
  long long ret = 0;
  while (true) {
    r = getchar();
    if ((r - '0' < 0 || r - '0' > 9) && r != '-' && !start) {
      continue;
    }
    if ((r - '0' < 0 || r - '0' > 9) && r != '-' && start) {
      break;
    }
    if (start) ret *= 10;
    start = true;
    if (r == '-')
      neg = true;
    else
      ret += r - '0';
  }
  if (!neg)
    return ret;
  else
    return -ret;
}
long long p[N][2], comp[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, q;
  cin >> n >> q;
  string st;
  cin >> st;
  p[0][0] = p[0][1] = 0;
  for (long long i = 0; i < st.size(); i++) {
    if (i != 0) {
      p[i][0] = p[i - 1][0];
      p[i][1] = p[i - 1][1];
    }
    long long cur = st[i] - '0';
    p[i][cur]++;
  }
  comp[0] = 1;
  for (long long i = 1; i < N; i++) {
    comp[i] = comp[i - 1] * 2;
    comp[i] %= mod;
  }
  while (q--) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    long long c0;
    if (l == 0)
      c0 = p[r][0];
    else
      c0 = p[r][0] - p[l - 1][0];
    long long ans = comp[r - l + 1] - comp[c0];
    ans += mod;
    ans %= mod;
    cout << ans << endl;
  }
}