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;
void smain();
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  smain();
}
vector<long long> pf;
const long long N = 1 << 17;
const long long M = 1e9 + 7;
long long cnt[N];
inline long long MOD(long long v) {
  if (v < 0) {
    v %= M;
    return v < 0 ? v + M : v;
  }
  return v % M;
}
long long prog(long long l, long long r) { return (l + r) * (r - l + 1) / 2; }
long long hlp[N] = {0, 1};
long long quer(long long l, long long r) {
  l--, r--;
  long long first = pf[r];
  if (l) first -= pf[l - 1];
  if (!first) return 0;
  long long ans = cnt[r - l + 1] - 1;
  long long v = (r - l + 1) - first;
  return MOD(ans - hlp[v]);
}
void smain() {
  cnt[0] = 1;
  for (long long i = 1; i < N; ++i) cnt[i] = (cnt[i - 1] * 2) % M;
  for (long long i = 2; i < N; ++i) hlp[i] = MOD(hlp[i - 1] + cnt[i - 1]);
  long long n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  pf.resize(n);
  for (long long i = 0; i < n; ++i) pf[i] = s[i] - '0';
  for (long long i = 1; i < n; ++i) pf[i] += pf[i - 1];
  for (long long i = 0; i < q; ++i) {
    long long l, r;
    cin >> l >> r;
    cout << quer(l, r) << '\n';
  }
}